42
Topic Page No. Theory 01 - 03 Exercise - 1 04 - 12 Exercise - 2 12 - 25 Exercise - 3 26 - 35 Exercise - 4 36 Answer Key 37 - 41 Contents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savart s law and Ampere s law ; Magnetic field near a current-carrying straight wire, along the axis of a circular coil and inside a long straight solenoid ; Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Magnetic moment of a current loop; Effect of a uniform magnetic field on a current loop; Moving coil galvanometer, voltmeter, ammeter and their conversions. Name : ____________________________ Contact No. __________________ ETOOSINDIA.COM India's No.1 Online Coaching for JEE Main & Advanced 3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

  • Upload
    others

  • View
    8

  • Download
    0

Embed Size (px)

Citation preview

Page 1: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

Topic Page No.

Theory 01 - 03

Exercise - 1 04 - 12

Exercise - 2 12 - 25

Exercise - 3 26 - 35

Exercise - 4 36

Answer Key 37 - 41

Contents

ELECTRO MAGNETIC FIELD (EMF)

SyllabusBiot-Savart�s law and Ampere�s law ; Magnetic field near a current-carrying

straight wire, along the axis of a circular coil and inside a long straight

solenoid ; Force on a moving charge and on a current-carrying wire in a

uniform magnetic field. Magnetic moment of a current loop; Effect of a

uniform magnetic field on a current loop; Moving coil galvanometer,

voltmeter, ammeter and their conversions.

Name : ____________________________ Contact No. __________________

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 2: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 1

1. A static charge produces only electric field and only electric field can exert a force on it.A moving charge produces both electric field ans magnetic field and both electric field and magnetic field canexert force on it.A current carrying conductor produces only magnetic field and only magnetic field can exert a force on it.

2. Magnetic charge (i.e. current) , produces a magnetic field . It can not produce electric field as net charge ona current carrying conductor is zero. A magnetic field is detected by its action on current carrying conductors

(or moving charges) and magnetic needles (compass) needles. The vector quantity B known as MAGNETIC

INDUCTION is introduced to characterise a magnetic field . It is a vector quantity which may be defined in termsof the force it produces on electric currents . Lines of magnetic induction may be drawn in the same way aslines of electric field . The number of lines per unit area crossing a small area perpendicular to the direction

of the induction bring numerically equal to B . The number of lines of

B crossing a given area is referred to

as the MAGNETIC FLUX linked with that area. For this reason B is also called MAGNETIC FLUX DENSITY .

3. MAGNETIC INDUCTION PRODUCED BY A CURRENT (BIOT-SAVART LAW):The magnetic induction dB produced by an element dl carrying a current I at a distance r is given by :

dB =

4r0

2r

sindI or

3

r0

r

rxdI

4dB

,

here the quantity Idl is called as current element strength. = permeability of the medium = 0 r ; 0 = permeability of free spacer

= relative permeability of the medium (Dimensionless quantity).Unit of 0 & is NA�2 or Hm�1 ; 0 = 4 × 10�7 Hm�1

4. MAGNETIC INDUCTION DUE TO A MOVING CHARGE :

20

P r4

sinqvdB

In vector form it can be written as 30

r

)rxv(q

4dB

5. MAGNETIC INDUCTION DUE TO AN INIFINITE ST. CONDUCTOR

B =

0

2

I

r6. MAGNETIC INDUCTION DUE TO SEMI INIFINITE ST. CONDUCTOR

B =

0I

r47. MAGNETIC INDUCTION DUE TO A CURRENT CARRYING STRAIGHT CONDUCTOR

B = R4

I0

(cos 1 + cos 2)

If the wire is very long 1 2 0º then , B =

R2

I0

8. MAGNETIC FIELD DUE TO A FLAT CIRCULAR COIL CARRYING A CURRENT :

(i) At its centre B = R2

IN0 , direction

Where N = total number of turns in the coilI = current in the coilR = Radius of the coil

(ii) On the axis B =

2/322

20

Rx2

RIN

Where x = distance of the point from the centre . It is maximum at the centre .

ELECTRO MAGNATIC FIELD

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 3: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 2

9. MAGNETIC INDUCTION DUE TO FLAT CIRCULAR ARC

B = R4

I

0

10. MAGNETIC INDUCTION DUE TO SOLENOID

B = 0nI, direction along axis.

where n no. of turns per m.

I current

11. MAGNETIC INDUCTION DUE TO TOROID

B = 0nI

where n = N

R2 (no. of turns per m)

N = total turns R >> r

12. MAGNETIC INDUCTION DUE TO CURRENT CARRYING SHEET

B = 1

20I

where I = Linear current density (A/m)

13. MAGNETIC INDUCTION DUE TO THICK SHEET

At point P2 Bout = 1

20Id

At point P1 Bin = 0Jx

14. GILBERT'S MAGNETISM (EARTH'S MAGNETIC FIELD) :(a) The line of earth's magnetic induction lies in a vetical plane coinciding with the magnetic North - South

direction at that place. This plane is called the MAGNETIC MERIDIAN. Earth's magnetic axis is slightly inclined tothe geometric axis of earth and this angle varies from 10.50 to 200. The Earth's Magnetic poles are oppositeto the geometric poles i.e. at earth's north pole, its magnetic south pole is situated and vice versa.

(b) On the magnetic meridian plane , the magnetic induction vector of the earth at any point, generally inclined

to the horizontal at an angle called the MAGNETIC DIP at that place , such that B = total magnetic induction of

the earth at that point.B v = the vertical component of

B in the magnetic meridian plane = B sin .

BH = the horizontal component of B in the magnetic meridian plane = B cos .

B

B

v

H

= tan .

(c) At a given place on the surface of the earth , the magnetic meridian and the geographic meridian maynot coincide . The angle between them is called "DECLINATION AT THAT PLACE" .

15. AMPERES LAWB d I.

I = algebric sum of all the currents .

16. LORENTZ FORCE :An electric charge 'q' moving with a velocity

V through a magnetic field of magnetic

induction B experiences a force

F , given by

F qVxB . There fore, if the charge moves

in a space where both electric and magnetic fields are superposed .F = nett electromagnetic force on the charge = q E qV B

This force is called the LORENTZ FORCE .

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 4: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 3

17. MOTION OF A CHARGE IN UNIFORM MAGNETIC FIELD :

(a) When v is || to

B : Motion will be in a st. line and

F = 0

(b) When v is | to

B : Motion will be in circular path with radius R = mv

qBand angular velocity =

qB

m and F = qvB.

(c) When v is at to

B : Motion will be helical with radius Rk =

mv

qB

sin and pitch

PH = 2 mv

qB

cos and F = qvBsin.

18. MAGNETIC FORCE ON A STRAIGHT CURRENT CARRYING WIRE : F I L B ( )

I = current in the straight conductorL = length of the conductor in the direction of the current in itB = magnetic induction. (Uniform throughout the length of conduction)

Note : In general force is

F I d B ( )

19. MAGNETIC INTERACTION FORCE BETWEEN TWO PARALLEL LONG STRAIGHT CURRENTS :When two long straight linear conductors are parallel and carry a current in each , theymagnetically interact with each other , one experiences a force. This force is of :

(i) Repulsion if the currents are anti-parallel (i.e. in opposite direction) or(ii) Attraction if the currents are parallel (i.e. in the same direction)

This force per unit length on either conductor is given by F =

0

2

I I

r1 2 . Where r = perpendicular distance

between the parallel conductors

20. MAGNETIC TORQUE ON A CLOSED CURRENT CIRCUIT :When a plane closed current circuit of 'N' turns and of area 'A' per turn carrying a currentI is placed in uniform magnetic field , it experience a zero nett force , but experience a

torque given by NIA B M B BINAsin

WhenA = area vector outward from the face of the circuit where the current is anticlockwise,

B = magnetic induction of the uniform magnetic feild.

M = magnetic moment of the current circuit = IN

A

Note : This expression can be used only if B is uniform otherwise calculus will be used.

21. MOVING COIL GALVANOMETER :It consists of a plane coil of many turns suspended in a radial magnetic feild. when a currentis passed in the coil it experiences a torque which produces a twist in the suspension. This deflection

is directly proportional to the torque NIAB = K

I = K

NAB

K = elastic torsional constant of the suspension

I = C C = K

NAB= GALVANOMETER CONSTANT..

22. FORCE EXPERIENCED BY A MAGNETIC DIPOLE IN A NON-UNIFORM MAGNETIC FIELD :

|F | = M

B

r

where M = Magnetic dipole moment.

23. FORCE ON A RANDOM SHAPED CONDUCTOR IN MAGNETIC FIELD

1. Magnetic force on a loop in a uniformB is zero

2. Force experienced by a wire of any shape is equivalent to force on a wire joiningpoints A & B in a uniform magnetic field .

24. MAGNETIC MOMENT OF A ROTATING CHARGE:If a charge q is rotating at an angular velocity ,

its equivalent current is given as I =q

2 & its

magnetic moment is M = IR2 = 1/2qR2.

NOTE : The rate of magnetic moment to Angular momentum of a uniform rotating object which is charged uniformlyis always a constant. Irrespective of the shape of conductor M/L = q/2m

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 5: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 4

PART - I : OBJECTIVE QUESTIONS

* Marked Questions are having more than one correct option.

SECTION (A) : MAGNET AND MAGNETIC FIELD DUE TO A MOVING CHARGE

A-1.* A magnetic needle (small magnet) is kept in a nonuniform magnetic field. It .

(A) may experience a force and torque (B) may experience a force but not a torque

(C) may experience a torque but not a force (D) will experience neither a force nor a torque

A-2. Two identical short magnetic dipoles of magnetic moments 1.0 A-m2 each, placed at a separation of 2m with their axes perpendicular to each other. The resultant magnetic field at a point midway betweenthe dipole is:

2m

(A) 5 × 10�7 T (B) 5 × 10�7 T (C) 10�7 T (D) 2 × 10�7 T

A-3. A point charge is moving in a circle with constant speed. Consider the magnetic field produced by the chargeat a fixed point P (not centre of the circle) on the axis of the circle.(A) it is constant in magnitude only(B) it is constant in direction only

(C) it is constant in direction and magnitude both(D) it is not constant in magnitude and direction both.

SECTION (B) : MAGNETIC FIELD DUE TO A STRAIGHT WIRE

B-1. Two infinitely long, thin, insulated, straight wires lie in the x-y plane along the x and y-axis respectively.Each wire carries a current I, respectively in the positive x-direction and positive y-direction. The magneticfield will be zero at all points on the straight line:

(A) y = x (B) y = � x (C) y = x � 1 (D) y = � x + 1

B-2. A current carrying wire is placed in the grooves of an insulating semi circular disc of radius 'R', asshown. The current enters at point A and leaves from point B. Determine the magnetic field at point D.

(A) 3R8

I0

(B)

3R4

I0

(C)

R4

I3 0

(D) none of these

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 6: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 5

B-3. Determine the magnitude of magnetic field at the centre of the current carrying wire arrangementshown in the figure. The arrangement extends to infinity. (The wires joining the successive squares arealong the line passing through the centre)

(A) a2

i0

(B) 0 (C)

a

i22 0

ln2 (D) none of these

SECTION (C) : MAGNETIC FIELD DUE TO A CIRCULAR LOOP, A STRAIGHT WIRE AND CIRCU-

LAR ARC, CYLINDER, LARGE SHEET, SOLENOID, TOROID AND AMPERE�S LAW

C-1. A current carrying wire AB of the length 2R is turned along a circle, as shown in figure. The magnetic fieldat the centre O.

O

A B

i

(A)

20

22

R2

i

(B)

22

R2

i0(C)

R2

i0 (2 � ) (D) R2

i0 (2 + )2

C-2. A battery is connected between two points A and B the circumference of a uniform conducting ring ofradius r and resistance R. One of the arcs AB of the ring subtends an angle at the centre. The valueof the magnetic induction at the centre due to the current in the ring is:

(A) zero, only if = 180º (B) zero for all values of (C) proportional to 2 (180º - ) (D) inversely proportional to r

C-3. A wire is wound on a long rod of material of relative permeability r = 4000 to make a solenoid. If the currentthrough the wire is 5 A and number of turns per unit length is 1000 per metre, then the magnetic field insidethe solenoid is :(A) 25.12 mT (B) 12.56 m T (C) 12.56 T (D) 25.12 T

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 7: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 6

C-4. A coaxial cable is made up of two conductors. The inner conductor is solid and is of radius R1 & theouter conductor is hollow of inner radius R2 and outer radius R3. The space between the conductors isfilled with air. The inner and outer conductors are carrying currents of equal magnitudes and in oppositedirections. Then the variation of magnetic field with distance from the axis is best plotted as:

(A) (B)

(C) (D)

C-5. Axis of a solid cylinder of infinite length and radius R lies along y-axis it carries a uniformly

distributed current � i � along +y direction. Magnetic field at a point

2R

,y,2R

is :-

(A) R4

i0

)k�i�( (B)

R2

i0

)k�j�( (C)

R4

i0

j� (D)

R4

i0

)k�i�(

C-6. Figure shows an amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is inhorizontal plane PQRS. Direction of circumlation along the path is shown by an arrow near point B and at D.

o d.B

for this path according to Ampere�s law will be :

(A) (i1 � i2 + i3) 0 (B) (� i1 + i2)0 (C) i30 (D) (i1 + i2)0

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 8: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 7

C-7. A cylindrical wire of radius R is carrying current i uniformly distributed over its cross-section. If a

circular loop of radius ' r ' is taken as amperian loop, then the variation value of dB over this loop

with radius ' r ' of loop will be best represented by:

(A) dB

R r

(B)

R r

dB

(C)

R r

dB (D)

R r

dB

SECTION (D) : MAGNETIC FORCE ON A CHARGE

D-1. Which of the following particles will experience minimum magnetic force (magnitude) when projected withthe same velocity perpendicular to a magnetic field?(A) Be +++ (B) proton (C) -particle (D) Li++

D-2. Electric current i enters and leaves a square loop made of homogeneous wire of uniform cross-sectionthrough diagonally opposite corners. A charge particle q moving along the axis of the square loop. Passesthrough centre at speed . The magnetic force acting on the particle when it passes through the centre hasa magnitude

(A) qa

0

2

i(B) q

a

0

2

i(C) q

a0i

(D) zero

D-3. Two particles X and Y having equal charges, after being accelerated through the same potential difference,enter a region of uniform magnetic field and describe circular paths of radii R

1 and R

2 respectively. The

ratio of the masses of X to that of Y.

(A)

2/1

2

1

R

R

(B)

1

2

R

R(C)

2

2

1

R

R

(D)

2

1

R

R

D-4. A negative charged particle falling freely under gravity enters a region having uniform horizontal mag-netic field pointing towards north. The particle will be deflected towards(A) East (B) West (C) North (D) South

D-5. A proton of mass m and charge q enters a magnetic field B with a velocity v at an angle with the directionof B. The radius of curvature of the resulting path is

(A) qBmv

(B) qBsinmv

(C) sinqB

mv(D) qB

cosmv

D-6. A current I flows along the length of an infinitely long, straight, thin walled pipe. Then(A) the magnetic field at all points inside the pipe is the same, but not zero(B) the magnetic field at any point inside the pipe is zero(C) the magnetic field is zero only on the axis of the pipe(D) the magnetic field is different at different points inside the pipe.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 9: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 8

D-7.* H+, He+ and O2+ all having the same kinetic energy pass through a thin region in which there is auniform magnetic field perpendicular to their velocity. The masses of H+, He+ and O2+ are 1 amu, 4amuand 16 amu respectively, then

(A) H+ will be deflected most

(B) O2+ will be deflected most

(C) He+ and O2+ will be deflected equally

(D) All will be deflected equally

D 8.* A beam of electrons moving with a momentum p enters a uniform magnetic field of flux density Bperpendicular to its motion. Which of the following statement(s) is (are) true?

(A) Energy gained is m2

p2

(B) Centripetal force on the electron is Be pm

(C) Radius of the electron's path is Bep

(D) Work done on the electrons by the magnetic field is zero

SECTION (E) : ELECTRIC AND MAGNETIC FORCE ON A CHARGE

E-1. A positively charged particle moves in a region having a uniform magnetic field and uniform electric field insame direction. At some instant, the velocity of the particle is perpendicular to the field direction. The path ofthe particle will be

(A) a straight line (B) a circle

(C) a helix with uniform pitch (D) a helix with increasing pitch.

E-2.* If a charged particle at rest experiences no electromagnetic force,

(A) the electric field must be zero (B) the magnetic field must be zero

(C) the electric field may or may not be zero (D) the magnetic field may or may not be zero

E-3.* If a charged particle projected in a gravity-free room it does not deflect,

(A) there must be an electric field (B) there may be a magnetic field

(C) both field cannot be zero (D) both fields can be nonzero

E-4.* Two ions have equal masses but one is singly-ionized and other is tripply-ionized. They are projected fromthe same place in a uniform magnetic field with the same velocity perpendicular to the field.

(A) Both ions will go along circles of equal radii.

(B) The circle described by the single-ionized charge will have a radius tripply that of the other circle

(C) The two circles do not touch each other

(D) The two circles touch each other

E-5.* A positively charged particle is moving along the positive X-axis. You want to apply a magnetic field for a shorttime so that the particle may reverse its direction and move parallel to the negative X-axis. This can be doneby applying the magnetic field along.

(A) Y-axis (B) Z-axis (C) Y-axis only (D) Z-axis only.

ETOOSINDIA.COM

India's No.1 Online Coaching for JEE Main & Advanced3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005

HelpDesk : Tel. 092142 33303

Page 10: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 9

SECTION (F) : MAGNETIC FORCE ON A CURRENT CARRYING WIRE

F-1. A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic fieldB such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is

(A) i r B (B) 2 r i B (C) zero (D) r i B

F-2. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallelto one of the sides of the loop and the plane of the loop is same of the left wire. If a steady current I isestablished in the wire as shown in the (fig) the loop will -

(A) Rotate about an axis parallel to the wire (B) Move away from the wire(C) Move towards the wire (D) Remain stationary.

F-3. A uniform magnetic field B

= k�j�4i�3 exists in region of space. A semicircular wire of radius 1 m

carrying current 1 A having its centre at (2, 2, 0) is placed in x-y plane as shown in fig. The force onsemicircular wire will be

(A) )k�j�i�(2 (B) )k�j�i�(2 (C) )k�j�i�(2 (D) )k�j�i�(2

F-4. Select the correct alternative(s):Two thin long parallel wires separated by a distance 'b' are carrying a current 'i' ampere each. Themagnitude of the force per unit length exerted by one wire on the other is

(A) 0

2

2

i

b(B)

02

2

i

b(C)

0

2

i

b(D)

022

i

b

F-5. In the figure shown a current 1 is established in the long straight wire AB. Another wire CD carrying current2 is placed in the plane of the paper. The line joining the ends of this wire is perpendicular to the wire AB. Theresultant force on the wire CD is:

(A) zero (B) towards negative x-axis(C) towards positive y-axis (D) none of these

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 11: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 10

SECTION (G) : MAGNETIC FORCE AND TORQUE ON A CURRENT CARRYING LOOP AND

MAGNETIC DIPOLE MOMENT

G-1. A bar magnet has a magnetic moment 2.5 JT�1 and is placed in a magnetic field of 0.2 T. Work done inturning the magnet from parallel to antiparallel position relative to the field direction.(A) 0.5 J (B) 1 J (C) 2.0 J (D) Zero

G-2. A circular loop of area 1 cm2, carrying a current of 10 A, is placed in a magnetic field of 0.1 T perpendicularto the plane of the loop. The torque on the loop due to the magnetic field is

(A) zero (B) 10-4 N-m (C) 10�2 N-m (D) 1 N-m

SECTION (H) : MAGNETIC FIELD DUE TO EARTH

H-1. A power line lies along the east-west direction and carries a current of 10 ampere. The force per metredue to the earth's magnetic field of 10�4 T is(A) 10�5 N (B) 10�4 N (C) 10�3 N (D) 10�2 N

H-2. A circular coil of radius 20 cm and 20 turns of wire is mounted vertically with its plane in magneticmeridian. A small magnetic needle (free to rotate about vertical axis) is placed at the center of the coil.It is deflected through 45° when a current is passed through the coil and in equilibrium (Horizontal

component of earth's field is 0.34 × 10�4 T). The current in coil is:

(A) 10

17 AA (B) 6A (C) 6 × 10�3 A (D)

503

AA

SECTION () : MISCELLENEOUS

-1. The magnetic materials having negative magnetic susceptibility are:(A) Non magnetic (B) Para magnetic (C) Diamagnetic (D) Ferromagnetic

PART - II : MISLLANEOUS QUESTIONS

Comprehension :

Curves in the graph shown give, as functions of radial distance r (from the axis), the magnitude B of themagnetic field (due to individual wire) inside and outside four long wires a, b, c and d, carrying currentsthat are uniformly distributed across the cross sections of the wires. Overlapping portions of the plotsare indicated by double labels. All curves start from the origin.

Ba

ba,c

b,dr

ca,b

c,d

1. Which wire has the greatest radius ?

(A) a (B) b (C) c (D) d

2. Which wire has the greatest magnitude of the magnetic field on the surface ?

(A) a (B) b (C) c (D) d

3. The current density in wire a is(A) greater than in wire c (B) less than in wire c.(C) equl to that in wire c. (D) not comparable to that of in wire c due to lack of information.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 12: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 11

Match The column

4. Column-II gives four situations in which three (in q,r,s) and four (in p) semi infinite current carrying wires areplaced in xy-plane as shown. The magnitude and direction of current is shown in each figure. Column-I givesstatements regarding the x and y components of magnetic field at a point P whose coordinates are P (0, 0,d). Match the statements in column-I with the corresponding figures in column-II and indicate your answerby darkening appropriate bubbles in the 4 × 4 matrix given in OMR.

Column-I Column-II

(A) The x component of magnetic field at (p) point P is zero in

(B) The z component of magnetic field at (q) point P is zero in

(C) The magnitude of magnetic field at (r)

point P is d4

i0

in

(D) The magnitude of magnetic field at (s)

point P is less than d2

i0

in

5. There are four situations given in column involving a magnetic dipole of dipole moment

placed in uniform

external magnetic field B

. Column gives corresponding results. Match the situtations in column with thecorresponding results in column

Column - Column -

(A) Magnetic dipole moment

is parallel (p) force on dipole is zero

to uniform external magnetic field B

(anglebetween both vectors is zero)

(B) Magnetic dipole moment , is perpendicular (q) torque on dipole is zero

to uniform external magnetic field B

(C) Angle between magnetic dipole moment

(r) magnitude of torque is (B)

and uniform external magnetic field B

is acute

(D) Angle between magnetic dipole moment

(s) potential energy of dipole due to

and uniform external magnetic field B

is 180o. external magnetic field is (B)

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 13: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 12

6. A particle enters a space where exists uniform magnetic field B B i B j B kx y z & uniform

electric field E E i E j E kx y z with initial velocity

u u i u j u kx y z . Depending on the values

of various components the particle selects a path. Match the entries of column A with the entriesof column B. The components other than specified in column A in each entry are non-zero. Neglectgravity.

Column-A Column-B

(i) By = B

z = E

x = E

z = 0 (P) circle

u = 0 (Q) helix with uniform pitchand constant radius

(ii) E = 0 ; uxB

x + u

yB

y u

zB

z(R) cycloid

(iii) uxB uxE 0 0, (S) helix with uniform pitch

and variable radius

(iv) u B B E , || (T) unknown curve

(U) helix with variable pitchand constant radius

(V) straight line

PART - I : MIXED OBJECTIVE

* Marked Questions are having more than one correct option.

Single choice type

1. A charge particle in the motion may produce(A) electric field only (B) magnetic filed only (C) both of them (D) none of these

2. Two parallel, long wires carry currents i1 and i

2 with i

1 > i

2. When the current are in the same direction, the

magnetic field at a point midway between the wire is 20T. If the direction of i1 is reversed, the field becomes

30T. The ratio i1/i

2 is

(A) 4 (B) 3 (C) 5 (D) 1

3. Consider a long, straight wire of cross-section area A carrying a current i. Let there be n free electrons perunit volume. An observer sitting in the car moving in the same direction to the current with a speed = (i/nAe)and separated from the wire by a distance r. The magnetic field seen by the observer is

(A)

0

2

i

r(B) zero (C)

0i

r(D)

2 0

i

r

4. A long, straight wire carries a current along the Z-axis. One can not find two points in the X-Y plane such that(A) the magnetic fields are equal in magnitude and same in direction(B) the directions of the magnetic fields are the same(C) the magnitudes of the magnetic fields are equal(D) the field at one point is opposite to that at the other point.

5. A vertical wire carries a current in downward direction. An electron beam sent horizontally towards the wirewill be deflected (gravity free space)

(A) towards right (B) towards left (C) upwards (D) downwards

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 14: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 13

6. A current-carrying, straight wire is kept along the axis of a square loop carrying a current. The straight wire(A) will exert an inward force on the square loop(B) will exert an outward force on the squareloop(C) will not exert any force on the square loop(D) will exert a force on the square loop parallel to itself.

7. A proton beam is going from west to east and an electron beam is going from east to west. Neglecting theearth�s magnetic field, the electron beam will be deflected

(A) towards the proton beam (B) away from the proton beam(C) away from the electron beam (D) None of these

8. A proton is moved along a magnetic field line. The magnetic force on the particle is(A) along its velocity (B) opposite to its velocity(C) perpendicular to its velocity (D) zero.

9. Two parallel wires carry currents of 10 A and 40 A in opposite directions. Another wire carrying a currentantiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be(A) towards 20 A (B) towards 40 A (C) zero(D) perpendicular to the plane of the currents

10. A toroid of mean radius ' a ' , cross section radius ' r ' and total number of turns N. It carries a current ' i '. Thetorque experienced by the toroid if a uniform magnetic field of strength B is applied :(A) is zero(B) is B i N r

2

(C) is B i N a2

(D) depends on the direction of magnetic field.

11. A long, thick straight conductor of radius R carries current uniformly distributed in its cross section area.The ratio of energy density of the magnetic field at distance R/2 from surface inside the conductor andoutside the conductor is:(A) 1: 16 (B) 1: 1 (C) 1: 4 (D) 9/16

12. A steady current 'l' flows in a small square loop of wire of side L in a horizontal plane. The loop is now

folded about its middle such that half of it lies in a vertical plane. Let 1

and 2

respectively denote the

magnetic moments of the current loop before and after folding. Then :

(A) 2

= 0

(B) 1

and 2

are in the same direction

(C) 22

1

(D) 2

1

2

1

13. A proton of mass 1.67 × 10�27 kg and charge 1.6 × 10�19 C is projected with a speed of 2 × 106 m/s at anangle of 60° to the x-axis. If a uniform magnetic field of 0.104 T is applied along the y-axis, the path of

the proton is :

(A) A circle of radius 0.2 m and time period × 10�7 s

(B) A circle of radius 0.1 m and time period 2 × 10�7 s

(C) A helix of radius 0.1 m and time period 2 × 10�7 s

(D) A helix of radius 0.2 m and time period 4 × 10�7 s

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 15: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 14

More than one choice type

14.* The magnetic field at the origin due to a current element i

d placed at a position r

is

(A)

0

4

i d xr

r

l

3(B)

0

4

i3r

dr

(C)

4i0

3r

dr

(D)

0

4

i3r

rd

15.* Consider three quantities x = E/B, y = 00

1

and z = CR. Here, is the length of a wire, C is a capaci-

tance and R is a resistance. All other symbols have standard meanings(A) x, y have the same dimensions. (B) y, z have the same dimensions(C) z, x have the same dimensions (D) one of the three pairs have the same dimensions

16.* A hollow tube is carrying an electric current along its length distributed uniformly over its surface. Themagnetic field(A) increases linearly from the axis to the surface(B) is constant inside the tube(C) is zero at the axis(D) is non-zero outside the tube at finite distance from surface

17.* In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in oppositedirections. The magnetic field is non-zero.

(A) outside the cable(B) inside the inner conductor except axis of the conductor(C) all the point inside the outer conductor(D) in between the two conductors.

18.* A particle of charge +q and mass m moving under the influence of a uniform electric field E i� and a

uniform magnetic field Bk� follows a trajectory from P and Q as shown in figure. The velocities at P and

Q are v i� and �2v j� . Which of the following statement(s) is/are correct?

a

v

2a

E

B

x

y

2v

P

Q

(A) E =

qamv

43 2

(B) Rate of work done by the electric field at P is

amv

43 3

(C) Rate of work done by the electric field at P is zero.

(D) Rate of work done by both fields at Q is zero.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 16: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 15

19.* Let E

and B

denote the electric and magnetic fields in a certain region of space. A proton moving with

a velocity v

along a straight line enters the region and is found to pass through it undeflected. Indicatewhich of the following statements are possible for the observation:

(A) E

= 0 and B

= 0

(B) 0E

and B

= 0

(C) 0E

, 0B

and both E

and B

are parallel to v

(D) E

is parallel to v

but B

is perpendicular to v

PART - II : SUBJECTIVE QUESTIONS

1. The magnetic moment of a short dipole is, 1 A m2. What is the magnitude of the magnetic inductionin air at 10 cm from centre of the dipole on a line making an angle of 30º from the axis of the dipole?

2. A point charge q = 2C is at the origin. It has velocity 2 i� m/s. Find the magnetic field at the following points

in vector form (at the moment when the charged particle passes through the origin) :(i) (2, 0, 0) (ii) (0, 2, 0) (iii) (0, 0, 2) (iv) (2, 1, 2)(v) Is the magnitude of the magnetic field on the circumference of the circle (in yz plane) y2 + z2 = c2 where �c�is a constant is same every where. Is it same in direction also.(vi) Answer the above (v) for the circle of same equation but in a plane x = a where �a� is a constant.

3. A particle of negative charge of magnitude �q� is revolving with constant speed ' V' in a circle of radius �R� as

shown in figure. Find the magnetic field (magnitude and direction) at the following points :

(i) centre of the circle (magnitude and direction)(ii) a point on the axis and at a distance �x� from the centre of the ring (magnitude only). Is its direction

constant all the time?

4. A pair of stationary and infinitely long bent wires is placed in the X-Y plane as shown in figure. Thewires carry currents of 10A each as shown. The segments L and M are along the x-axis. The segmentsP and Q are parallel to the Y - axis such that OS = OR = 0.02 m. Find the magnitude and directionof the magnetic induction at the origin O.

5. A current of 1 amp is flowing in the sides of an equilateral triangle of side 4.5 x 10 -2 m. Find themagnetic field at the centroid of triangle.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 17: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 16

6. Two straight infinitely long and thin parallel wires are spaced 0.1 m apart and carry a current of 10ampere each. Find the magnetic field at a point distant 0.1 m from both wires in the two cases whenthe currents are in the(i) Same and(ii) Opposite direction.

7. Four infinitely long 'L' shaped wires, each carrying a current i have been arranged as shown in thefigure. Obtain the magnetic field strength at the point 'O' equidistant from all the four corners.

8. Figures shows a long wire bent at the middle to form a right angle. Show that the magnitudes of the magneticfields at the points Q and R are unequal and find these magnitudes. The wire w

1 and the circumference of

circle are coplaner and w2 is perpendicular to plane of paper. Also find the ratio of field at Q and R

9. A long wire carrying a current i is bent to form a plane angle . Find the magnetic field B at a point on the

bisector of this angle situated at a distance x from the vertex is written in the form of 4

cotK

Tesla. Then, find

the value of K.

10. Find the magnetic field B at the centre of a square loop of side 'a', carrying a current i.

11. Each of the batteries shown in figuer has an emf equal to 10 V. Find the magnetic field B at the point p.

P

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 18: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 17

12. (i) Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 1 A. The coils have 50 and 100turns respectively and are placed in such a way that their planes as well as the centre coincide. Findthe magnitude of the magnetic field B at the common centre when the currents in the coils are (a) inthe same sense (b) in the opposite sense.

(ii) If the outer coil of the above problem is rotated through 90º about a diameter, what would be the

magnitude of the magnetic field B at the centre?

13. Two circular coils of wire each having a radius of 4 cm and 10 turns have a common axis and are 6 cmapart. If a current of 1 A passes through each coil in the opposite direction find the magnetic induction.(a) At the centre of each coil ;(b) At a point on the axis, midway between them.

14. Two wire loops PQRSP formed by joining two semicircular wires of radii R1 and R

2 carries a current I as

shown in (fig.) The magnitude of the magnetic induction at the center C is.....

R2 R1

I

I

Q PS R C

15. Find the magnitude of the magnetic induction B of a magnetic field generated by a system of thinconductors (along which a current i is flowing) at a point A (0, R, 0), that is the centre of a circularconductor of radius R. The circular part is in yz plane.

16. Find the magnetic induction of the field at the point O of a loop with current , whose shape isillustrated in figure

(a) In figure 'a' the radii a and b, as well as the angle are known,(b) In figure b, the radius a and the side b are known.(c) A current = 5.0 A flows along a thin wire shaped as shown in figure. The radius of a curved part

of the wire is equal to R = 120 mm, the angle 2 = 90°. Find the magnetic induction of the field

at the point O.

0

2R

2

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 19: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 18

17. Find the magnetic induction at the point O if the wire carrying a current has the shape shown in figurea, b, c. The radius of the curved part of the wire is R, the linear parts of the wire are very long.

z

x

y0R

z

x

y0R

z

x

y0R

(a) (b) (c)

18. A conductor consists of an infinite number of adjacent wires, each infinitely long & carrying a current i.Show that the lines of B will be as represented in figure & that B for all points in front of the infinitecurrent sheet will be given by, B = (1/2)0 ni, where n is the number of conductors per unit length.

19. Figure shows a cylindrical conductor of inner radius a & outer radius b which carries a current i uni-formly spread over its cross section. Show that the magnetic field B for points inside the body of the

conductor (i.e. a < r < b) is given by, B = rar

)ab(2

i 22

220

. Check this formula for the limiting case

of a = 0.

20. A thin but long, hollow, cylindrical tube of radius r carries a current i along its length. Find the magnitude ofthe magnetic field at a distance r/4 from the surface (a) inside the tube (b) outside the tube.

21. The magnetic field B inside a long solenoid, carrying a current of 10 A, is 3.14 × 10�2 T. Find the number ofturns per unit length of the solenoid.

22. A copper wire having resistance 0.01 ohm in each metre is used to wind a 400 turn solenoid of radius 1.0 cmand length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magneticfield of 1.0 × 10�2 T near the centre of the solenoid.

23. A charged particle is accelerated through a potential difference of 24 kV and acquires a speed of 2×106 m/s.It is then injected perpendicularly into a magnetic field of strength 0.2 T. Find the radius of the circle describedby it.

24. A neutron, a proton, an electron and an -particle enters a uniform magnetic field with equal velocities.The field is directed along the inward normal to the plane of the paper. Which of these tracks followedare by electron and -particle.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 20: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 19

25. In the formula X = 3 YZ2, the quantities X and Z have the dimensions of capacitance and magneticinduction respectively. The dimensions of Y in the MKS system are..............

26. Two long parallel wires carrying currents 2.5 amps and I amps in the same direction (directed into theplane of the paper) are held at P and Q respectively such that they are perpendicular to the plane ofpaper. The points P and Q are located at a distance of 5m and 2m respectively from a collinear point R.

(a) An electron moving with a velocity of 4 x 105 m/s along the positive X-directionexperiences a force of magnitude 3.2 x 1020 N at the point R. Find the value of I.

(b) Find all the positions at which a third long-parallel wire carrying a current of magnitude2.5 A may be placed so that the magnetic induction at R is zero.

27. An particle is accelerated by a potential difference of 104V. Find the change in its direction of motion,if it enters normally in a region of thickness 0.1 m having transverse magnetic induction of 0.1 Tesla.(Given: mass of -particle is equal to 6.4 × 10�27 kg)

28. A magnetic field of 8 k� mT exerts a force of (4.0 i� + 3.0 j� ) × 10�10 N on a particle having a charge of

5 × 10�10 C and going in the X � Y plane. Find the velocity of the particle.

29. An experimenter�s diary reads as follows; �a charged particle is projected in a magnetic field of

(7.0 i� � 3.0 j� ) × 10�3 T. The acceleration of the particle is found to be (x i� + 7.0 j� ) × 10�6 m/s2. Find the value

of x.

30. A particle having a charge of 2.0 × 10�8 C and a mass of 2.0 × 10�10 g is projected with a speed of 2.0 × 103

m/s in a region having a uniform magnetic field (B = 0.1 T). Find the radius of the circle formed by the particleand also the frequency.

31. A proton describes a circle of radius 1 cm in a magnetic field of strength 0.10 T. What would be the radius ofthe circle described by an deuterium moving with the same speed in the same magnetic field?

32. An electron having a kinetic energy of 400 eV circulates in a path of radius 20 cm in a magnetic field. Find themagnetic field and the number of revolutions per second made by the electron.

33. A proton is projected with a velocity of 3 × 106 m/s perpendicular to a uniform magnetic field of 0.6T. Find the

acceleration of the proton. mass of proton =2710

35 kg

34. (a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0.50 T.What would be its speed? Is it reasonable?

(b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its

speed? mass of proton = 271035 kg

35. A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B asshown in figure. (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the anglesubtended by the arc at the centre. (c) How long does the particle stay inside the magnetic field? (d) Solvethe three parts of the above problem if the charge q on the particle is negative.

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

v

B

/4

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 21: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 20

36. A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find theangle of deviation (figure) of the particle as it comes out of the magnetic field if the width d of the region is veryslightly smaller than

) × × × × ×

× × × × ×

× × × × ×B

d

(a) qBmv

(b) qB2

mv(c) qB

mv3

37. Figure shows a convex lens of focal length 10 cm lying in a uniform magnetic field B of magnitude 1.2 Tparallel to its principal axis. A particle having a charge 2.0 × 10�3 C and mass 2.0 × 10�5 kg is projectedperpendicular to the plane of the diagram with a speed of 4.8 m/s. The particle moves along a circle with itscentre on the principal axis at a distance of 15 cm from the lens. The axis of the lens and of the circle aresame. Show that the image of the particle goes along a circle and find the radius of that circle.

B

P

38. A particle having a charge of 5.0 C and a mass of 5.0 × 10�12 kg is projected with a speed of 1.0 km/sin a magnetic field of magnitude 5.0 mT. The angle between the magnetic field vector and the velocity vectoris sin�1 (0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

39. A proton projected in a magnetic field of 0.04 T travels along a helical path of radius 5.0 cm and pitch 20 cm.Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take themass of the proton = 1.6 × 10�27 kg.

40. An electron beam passes through a magnetic field of 2 × 10�3 Wb/m2 and an electric field of 3.2 × 104 V/m,

both acting simultaneously. ( VBE

) If the path of electrons remains undeflected calculate the speed ofthe electron. If the electric field is removed, what will be the radius of the electron path[mass of electron = 9.1 × 10�31 kg]?

41. If two charged particles of same mass and charge are describing circles in the same magnetic fieldwith radii r

1 and r

2 (> r

1), the speed of the first particle is.... that of the second particle while the time

period of the particle is...... that of the second particle.

42. A conducting wire of length , lying normal to a magnetic field B, moves with a velocity v as shown in figure.(a) Find the average magnetic force on a free electron of the wire. (b) Due to this magnetic force, electronsconcentrate at one end resulting in an electric field inside the wire. The redistribution stops when the electricforce on the free electrons balances the magnetic force. Find the electric field developed inside the wire whenthe redistribution stops. (c) What potential difference is developed between the ends of the wire?

× × × ×

× × ×

× × × ×

v

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 22: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 21

43. A current i is passed through a cylindrical gold strip of radius r. The number of free electrons per unit volumeis n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region as shown infigure, what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the freeelectrons get accumulated on one side of the conductor along its length. This produces a transverse electricfield in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electricfield which will stop further accumulation of electrons. (d) What will be the potential difference developedacross the width of the conductor due to the electron accumulation? The appearance of a transverse emf,when a current-carrying wire is placed in a magnetic field, is called Hall effect.

× × × ×

× × ×

× × × ×

×

×

×

×

×

×

× × × × × ×

i

44. A uniform magnetic field of magnitude 0.20 T exists in space from east to west. A particle having mass 10�5

kg and charge 10�5 C is projected from south to north so that it moves with a uniform velocity. Find velocity ofprojection of the particle? (g = 10 m/s2)

45. A particle moves in a circle of radius 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of200 V/m makes the path straight. Find the charge/mass ratio of the particle.

46. A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to eachother) at a speed of 10 5 m/s. The velocity is perpendicular to both the fields. When the electric field isswitched off, the proton moves along a circle of radius 2 cm. Find the magnitudes of the electric andthe magnetic fields. Take the mass of the proton = 1.6 × 10�27 kg.

47. A particle having mass m and charge q is released from the origin in a region in which electric field andmagnetic field are given by

j�BB 0

and i�EE 0

.

Find the speed of the particle as a function of its X-coordinate.

48. Consider a 10 cm long portion of a straight wire carrying a current of 10 A placed in a magnetic field of 0.1 Tmaking an angle of 37º with the wire. What magnetic force does the wire experience?

49. A current of 2 A enters at the corner d of a square frame abcd of side 10 cm and leaves at the opposite cornerb. A magnetic field B = 0.1 T exists in the space in direction perpendicular to the plane of the frame as shownin figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.

d c

a b

50. A magnetic field of strength 1.0 T is produced by a strong electromagnet in a cylindrical region of diameter4.0 cm as shown in figure. A wire, carrying a current of 2.0 A, is placed perpendicular to and intersecting theaxis of the cylindrical region. Find the magnitude of the force acting on the wire.

S

Ni

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 23: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 22

51. A wire of length l carries a current i along the y-axis. A magnetic field exists which is given as T)k�j�i�(BB 0

.

Find the magnitude of the magnetic force acting on the wire.52. A current of 10 A exists in the circuit shown in figure. The wire PQ has a lengths of 100 cm and the magnetic

field in which it is immersed has a magnitude of 0.20 T. Find the magnetic force acting on the wire PQ.

× × × × ×

× × × × ×

× × × × ×QP

53. A thin straight horizontal wire of length 0.2 m whose mass is 10�4 kg floats in a magnetic induction fieldwhen a current of 10 ampere is passed through it. To make this possible, what should be the minimummagnetic strength? (Take g = 10 m/s2)

54. Two long straight parallel conductors are separated by a distance of r1 = 5cm and carry constantcurrents i1 = 10 A & i2 = 20 A. What work per unit length of a conductor must be done to increase theseparation between the conductors to r2 = 10 cm if, currents flow in the same direction?

55. A wire, carrying a current, i, is kept in the X � Y plane along the curve y = A sin

x

2. A uniform magnetic

field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire betweenx = 0 and x = /2

56. A rigid wire consists of a semicircular portion of radius R and two straight sections (figure). The wire ispartially immersed in a perpendicular magnetic field B as shown in the figure. Find the magnetic force on thewire if it carries a current i.

57. A metal wire PQ of mass 10 gm lies at rest on two horizontal metal rails separated by 4.90 cm (figure). Avertically downward magnetic field of magnitude 0.800 T exists in the space. The resistance of the circuit isslowly decreased and it is found that when the resistance goes below 20.0 , the wire PQ starts sliding onthe rails. Find the coefficient of friction. Neglect magnetic force acting on wire PQ due to metal rails (g = 9.8m/s2)

P

Q

× × × ×

× × × ×

× × × ×

× × × ×

6 V

58. The magnetic field existing in a region is given by

k�x

�1BB 0

, where B

0 and are constants, X is the X coordinate of a point and k� is the unit vector along

Z axis.A square loop of edge and carrying a current i, is placed with its edges parallel to the X, Y axes. Find themagnitude of the net magnetic force experienced by the loop.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 24: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 23

59. Two parallel wires separated by a distance of 10 cm carry currents of 20 A and 80 A along the same direction.Where should a third current carrying wire be placed so that it experiences no magnetic force?

60. Figure shows a part of an electric circuit. The wires AB, CD and EF are very long and have identical resis-tances. The separation between the neighboring wires is 2 cm. The wires AE and BF have negligible resis-tance and the ammeter reads 60 A. Calculate the magnetic force per unit length on AB and CD.

AC D

F

BA

E

61. A straight, long wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If theforce acting on unit length of the second wire is 2.0 × 10�4 N, what is the separation between them?

62. A circular coil of 100 turns has an effective radius 0.05 m and carries a current of 0.1 amp. How muchwork is required to turn it in an external magnetic field of 1.5 wb/m2 through 1800 about an axis perpen-dicular to the magnetic field. The plane of the coil is initially perpendicular to the magnetic field.

63. (a) A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. Thecentre of the loop coincides with the centre of the field (figure). The strength of the magnetic field atthe periphery of the loop is B. Find the magnetic force on the wire.

Bia

(b) A hypothetical magnetic field existing in a region is given by r0eBB

, where re

denotes the unit

vector along the radial direction of a point relative to the origin and B0 = constant. A circular loop of

radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the centre at (0,0, a). Find the magnitude of the magnetic force acting on the loop.

64. A rectangular coil of 100 turns has length 4 cm and width 5 cm. It is placed with its plane parallel to a uniformmagnetic field and a current of 2A is sent through the coil. Find the magnitude of the magnetic field B, if thetorque acting on the coil is 0.2 N-m.

65. A 50-turn circular coil of radius 4 cm carrying a current of 2.5 A is rotated in a magnetic field of strength 0.20T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torqueacting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

66. A square loop of sides 10 cm carries a current of 10 A. A uniform magnetic field of magnitude 0.20 T existsparallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque actingon the loop?

67. A circular coil of diameter 2.0 cm has 500 turns in it and carries a current of 1.0 A. Its axis makes an angleof 30º with the uniform magnetic field of magnitude 0.40 T that exists in the space. Find the torque acting on

the coil.

68. A circular loop carrying a current i has wire of total length L. A uniform magnetic field B exists parallel to theplane of the loop. (a) Find the torque on the loop. (b) If the same length of the wire is used to form arectangular loop of side ratio 1 : 2 , what would be the torque? Which is larger?

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 25: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 24

69. In a hydrogen atom the electron moves in an orbit of radius 0.5 Å making 1016 rev/s. What is themagnetic moment associated with the orbital motion of the electron and the magnetic field at thecentre?

70. A charge Q is spread uniformly over an insulated ring of radius R. What is the magnetic moment of thering if it is rotated with an angular velocity about its axis?

71. Two circular coils each of 100 turns are held such that one lies in the vertical plane and the other in thehorizontal plane with their centres coinciding. The radius of the vertical and the horizontal coils arerespectively 20 cm and 30 cm. If the directions of the current in them are such that the earth'smagnetic field at the centre of the coil is exactly neutralized, calculate the current in each coil.[horizontal component of the earth's field = 27.8 A m -1 (Tesla) ; angle of dip = 30º]

72. A short magnet of magnetic moment 6 Amp.m2 is lying in a horizontal plane with its North polepointing 60º East of North. Find the net horizontal magnetic field at a point on the axis of the magnet

0.2 m away from it. [ Horizontal component of earth's magnetic field = 0.3 x 10 -4 tesla ]

EW

N

S

S

N60°

73. A coil of 50 turns and 20 cm diameter is made with a wire of 0.2 mm diameter and resistivity 2×106

cm. The coil is connected to a source of EMF. 20 V and negligible internal resistance.

(a) Find the current through the coil .

(b) What must be the potential difference across the coil so as to nullify the earth's horizontalmagnetic induction of 3.14 × 10 -5 tesla at the centre of the coil. How should the coil be placedto achieve the above result.

74. A charge of 1 coulomb is placed at one end of a non-conducting rod of length 0.6m. The rod is rotatedin a vertical plane about a horizontal axis passing through the other end of the rod with an angularvelocity 104 rad/sec. Find the magnetic field at a point on the axis of rotation at a distance of 0.8mfrom the centre of the path. Now half of the charge is removed from one end and placed on the otherend. The rod is rotated in a horizontal plane about vertical axis passing through the mid point of the rodwith the same angular velocity. Calculate the magnetic field at the point on the axis at a distance of0.4m from the centre of the rod.

75. A square loop of wire of edge a carries a current i. Show that the value of B at the center is given by,

B =a

i22 0

.

76. A circular loop of radius r carries a current i. How should a long, straight wire carrying a current 3i be placedin the plane of the circle so that the magnetic field at the centre becomes zero?

77. A constant direct current of uniform density j

is flowing in an infinitely long cylindrical conductor. The

conductor contains an infinitely long cylindrical cavity whose axis is parallel to that of the conductor

and is at a distance

from it. Determine the magnetic induction B

inside the cavity..

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 26: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 25

78. A solenoid of length 0.4 m and diameter 0.6 m consists of a single layer of 1000 turns of fine wirecarrying a current of 5.0 × 10�3 ampere. Calculate the magnetic field strength on the axis at the middleand at the ends of the solenoid.

79. A charged particle +q of mass m is placed at a distance d from another charge particle 2q of mass 2m in a uniform magnetic field of induction vector B as shown in the fig. If the particles are projectedtowards each other with equal speeds v.

(a) Find the maximum value of the projection speed Vmax so that the two particles do not collide.

(b) Find the time interval after which collision occurs between the particles if projection speedequals 2Vmax.

(c) Assuming that the particles stick after the collision find the radius of the circular path of theparticle in subsequent motion. (Neglect the interaction between the particles)

80. A straight rod of mass m and length l can slide on two parallel plastic rails kept in a horizontal planewith a separation d. The coefficient of friction between the wire and the rails is . If the wire carries acurrent i, what minimum magnetic field should exist in the space in order to slide the wire on the rails.

81. A finite conductor AB carrying current i is placed near a fixed very long wire current carrying i0 asshown in the figure. Find the point of application and magnitude of the net ampere force on the conductorAB. What happens to the conductor AB if it is free to move. (Neglect gravitational field)

82. Figure shows (only cross section) a wooden cylinder C with a mass m of 0.25 kg, a radius R and alength perpendicular to the plane of paper of 0.1 meter with N = 10 where N is number of turns of wirewrapped around it longitudinally, so that the plane of the wire loop contains the axis of the cylinder.What is the least current through the loop that will prevent the cylinder from moving down a planewhose surface is inclined at angle to the horizontal, in the presence of a vertical field of magneticinduction 0.5 weber/meter2, if the plane of the windings is parallel to the inclined plane? (bottom mostpoint of cylinder does not slip)

B

C

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 27: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 26

PART-I IIT-JEE (PREVIOUS YEARS PROBLEMS)

*Marked Questions are having more than one correct option.

1. A current carrying loop is placed in a uniform magnetic field towards right in four different orientations,V, arrange them in the decreasing order of Potential Energy.[JEE 2003 (Screening) 3/84 ]

(i) (ii) (iii) (iv)

(A) , , V (B) , , , V (C) , V, , (D) , V, ,

2. A conducting loop carrying a current is placed in a uniform magnetic field pointing into the plane of thepaper as shown. The loop will have a tendency to. [JEE 2003 (Screening) 3/84 ]

(A) move along the positive x direction

(B) move along the negative x direction

(C) contract

(D) expand

3. A proton and particle, after accelerating through same potential difference enters into a uniform magnetic

field perpendicular to their velocities, find the radius ratio of proton and particle. [JEE 2004 (Mains), 2/60]

4. An electron traveling with a speed u along the positive x-axis enters into a region of magnetic field where

B = �B0 k� (x > 0). It comes out of the region with speed v then [JEE 2004 (Screening) 3/84 ]

(A) v = u at y > 0 (B) v = u at y < 0

(C) v > u at y > 0 (D) v > u at y < 0

5. Relation for a Galvanometer having number of turns N, area of cross section A and moment of inertia is

given as : = Ki where K is a positive constant and � i � is current in the coil placed in the magnetic field B.

[JEE 2005 Mains, 2+2+2/60](i) Find K in terms of B, N and A

(ii) Find torsional constant of spring if a current 0 produces a deflection of

2

(iii) If at an instant charge Q is flown through the galvanometer, find the maximum deflection in

the coil. (assume I as the moment of inertia of the coil )

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 28: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 27

6.* Which of the following statement is correct in the given figure. [JEE 2006 ; +5,-1/35]

(A) net force on the loop is zero

(B) net torque on the loop is zero

(C) loop will rotate clockwise about axis OO� when seen from O

(D) loop will rotate anticlockwise about OO� when seen from O

7. A magnetic field j�BB 0

exists in the region a < x < 2a and j�B�B 0

, in the region 2a < x < 3a, where B0 is a

positive constant. A positive point charge moving with a velocity i�vV 0

, where v0 is a positive constant, enters the

magnetic field at x = a. The trajectory of the charge in this region can be like. [JEE - 2007' +3, -1/162]

-B0

0 a 2a 3ax

B0

(A) 0a 2a 3a

x

z

(B) 0a 2a 3a

x

z

(C) 0a 2a 3a

x

z

(D) 0a 2a 3a

x

z

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 29: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 28

8. Two wires each carrying a steady current are shown in four configurations in Column . Some of theresulting effects are described in Column . Match the statements in Column with the statements inColumn and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.

[JEE - 2007' 6/162]Column Column

(A) Point P is situated midway (p) The magnetic fields (B) at P due to the

between the wires. P currents in the wires are in the same

direction.(B) Point P is situated at the mid�point of the line (q) The magnetic field B at P due to the

joining the centres of the circular wires, currents in the wires are in opposite

which have same radii. P

directions.

(C) Point P is situated at the mid�point of the line (r) There is no magnetic field at P.joining the centers of the circular wires,

which have same radii.P

(D) Point P is situated at the common center (s) The wires repel each other.

of the wires. P

9. A particle of mass m and charge q, moving with velocity V enters region normal to the boundary as shownin the figure. Region has a uniform magnetic field B perpendicular to the plane of the paper. The length ofthe region is . Choose the correct choice(s). [JEE - 2008' 4/163]

(A) The particle enters Region only if its velocity V > m

Bq

(B) The particle enters Region only if its velocity V < m

Bq

(C) Path length of the particle in Region is maximum when velocity V = m

Bq

(D) Time spent in Region is same for any velocity V as long as the particle returns to Region

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 30: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 29

10. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column�II.

In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be theelectric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now,the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be themagnetic field at M and be the magnetic moment of the system in this condition. Assume each rotatingcharge to be equivalent to a steady current. [JEE - 2009, 8/160 conducted by IIT Guwahati]

Column�I Column�II

(A) E = 0 (p) Charges are at the corners of aregular hexagon. M is at thecentre of the hexagon. PQ isperpendicular to the plane ofthe hexagon.

(B) V 0 (q)P

Q

M

� + � + � +

Charges are on a line perpendicular toPQ at equal intervals. M is themidpoint between the two innermostcharges.

(C) B = 0 (r) +� +

��

+P

Q

M

Charges are placed on two coplanarinsulating rings at equal intervals. M isthe common centre of the rings. PQ isperpendicular to the plane of rings.

(D) 0 (s) � �+

� �+

P QM

Charges are placed at the corners of arectangle of sides a and 2a and at themid�points of the longer sides. M is at

the centre of the rectangular. PQ isparallel to the longer sides.

(t) +

+ + M

P

Q

� �

Charges are placed on two coplanar,identical insulating rings at equalintervals. M is the mid�point

between the centres of the rings.PQ is perpendicular to the linejoining the centres and coplanar tothe rings.

11. A steady current goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR =

4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is

x48k 0

, find the value of k.

[JEE 2009, 4/160, �1]

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 31: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 30

12. A thin flexible wire of length L is connected to two adjacent fixed points carries a current in the clockwisedirection, as shown in the figure. When system is put in a uniform magnetic field of strength B going into theplane of paper, the wire takes the shape of a circle. The tension in the wire is : [JEE 2010, 3/163, �1]

(A) BL (B)

BL(C)

2BL

(D)

4BL

13. An electron and a proton are moving straight parallel paths with same velocity. They enter a semi-infiniteregion of uniform magnetic field perpendicular to the velocity. Which of the following statement(s) is/are true?

(A) They will never come out of the magnetic field region [IIT-JEE 2011; 4/160]

(B) They will come out travelling along parallel paths

(C) They will come out at the same time

(D) They will come out at different times.

14. A long insulated copper wire is closely wound as a spiral of �N� turns. The spiral has inner radius �a� and

outer radius �b�. The spiral lies in the X-Y plane and a steady current �I� flows through the wire. The Z-

component of the magnetic field at the center of the spiral [IIT-JEE 2011; 3/160]

(A) 0N

2(b a)

In

ba

(B) 0N

2(b a)

In

b ab a

(C) 0N

2b

In

ba

(D) 0N

2b

In

b ab a

15. Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and

magnetic fields 0E E j

and 0B B j

. At time t = 0, this charge has velocity in the x-y plane, making an

angle with the x-axis . Which of the following option(s) is(are) correct for time t > 0 ?

(A) If = 0º , the charge moves in a circular path in the x-z plane. [IIT-JEE 2012; 4/136]

(B) If = 0º, the charge undergoes helical motion with constant pitch along the y-axis.

(C) If = 10º the charge undergoes helical motion with its pitch increasing with time along the y-axis.

(D) If = 90º, the charge undergoes linear but accelerated motion along the y-axis.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 32: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 31

16. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a as shown in the figure. Both thecylinder and the cavity are infinitely long. A uniform current density J flows along the length. If the magnitude of

the magnetic field at the point P is given by N12

µ0aJ, then the value of N is ? [IIT-JEE 2012; 4/136]

17. A loop carrying current I lies in x-y plane as shown in the figure. The unit vector �k is coming out of the planeof the paper. The magnetic moment of the current loop is [IIT-JEE 2012; 3/136, � 1]

(A) a2 I �k (B) 12

a2 I �k

(C) � 12

a2 I �k (D) (2+1)a2 I �k

18. An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a unifrom current

density along its length. The magnitude of the magnetic field, B

as a function of the radial distance r from

the axis is best represented by [IIT-JEE 2012; 3/136, � 1]

(A) (B)

(C) (D)

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 33: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 32

19. A particle of mass M and positive charge Q, moving with a constant velocity

�11u 4ims , enters a region of

uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = Lfor all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds

with a velocity

2u 2 3i j ms�1. The correct statement(s) is (are) [JEE Advanced (P-1) 2013]

(A) The direction of the magnetic field is �z direction

(B) The direction of the magnetic field is +z direction

(C) The magnitude of the magnetic field 50 M

3Q units

(D) The magnitude of the magnetic field is 100 M

3Q units

20. A steady current I flows along an infinitely long hollow cylindrical conductor of radius R. This cylinder isplaced coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carriesa steady current I. Consider a point P at a distance r form the common axis. The correct statement (s) is(are) [JEE Advanced (P-2) 2013](A) In the region 0 < r < R, the magnetic field is non-zero.(B) In the region R < r < 2R, the magnetic field is along the common axis.(C) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis.(D) In the region r > 2R, the magnetic field is non-zero.

PART-II AIEEE (PREVIOUS YEARS PROBLEMS)

* Marked Questions are having more than one correct option.

1. A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60º. The torque

needed to maintain the needle in this position will be : [AIEEE 4/300 2003]

(1) W3 (2) W (3) W)2/3( (4) 2 W

2. A magnetic lines of force inside a bar magnet : [AIEEE 4/300 2003](1) are from north-pole to south-pole of the magnet(2) do not exist

(3) depend upon the area of cross-section of the bar magnet(4) are from south-pole to north-pole of the magnet

3. A current i ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at anypoint inside the tube is : [AIEEE 4/300 2004]

(1) infinite vuUr (2) zero 'kwU; (3) ri2

,4

0

tesla (4)

ri2 tesla

4. A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre ofthe coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be:

[AIEEE 4/300 2004](1) nB (2) n2B (3) 2nB (4) 2n2B

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 34: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 33

5. The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cmfrom the centre is 54 T. What will be its value at the centre of the loop ? [AIEEE 4/300 2004](1) 250 T (2) 150 T (3) 125 T (4) 75 T

6. Two long conductors, separated by a distance d carry currents I1 and I

2 in the same direction. They exert a force

F on each other. Now the current in one of them is increased to two times and its direction is reversed. Thedistance is also increased to 3d. The new value of the force between them is : [AIEEE 4/300 2004](1) �2 F (2) F/3 (3) �2F/3 (4) � F/3

7. The length of a magnet is large compared to its width and breadth. The time period of its oscillation in avibration magnetometer is 2s. The magnet is cut along its length into three equal parts and three parts arethen placed on each other with their like poles together. The time period of this combination will be :

[AIEEE 4/300 2004]

(1) 2s (2) 2/3 s (3) s 32 (4) s 3/2

8. The materials suitable for making electromagnets should have : [AIEEE 4/300 2004](1) high retentivity and high coercivity (2) low retentivity and low coercivity(3) high retentivity and low coercivity (4) low retentivity and high coercivity

9. Two thin, long, parallel wires, separated by a distance �d� carry a current of �i� A in the same direction. They

will :

(1) attract each other with a force of )d2(

i20

(2) repel, each other with a force of

)d2(

i20

(3) attract each other with a force of )d2(

i2

20

(4) repel each other with a force of

)d2(

i2

20

10. Two concentric coils each of radius equal to 2 cm are placed at right angles to each other. 3 ampere and 4ampere are the currents flowing in each coil respectively. [AIEEE 4/300 2005]The magnetic induction in weber/m2 at the centre of the coils will be (

0 = 4p × 10�7 Wb/A.m):

(0 = 4p × 10�7 Wb/A.m):

(1) 12 × 10�5 (2) 10�5 (3) 5 × 10�5 (4) 7 × 10�5

11. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If anelectron is projected along the direction of the fields with a certain velocity, then : [AIEEE 4/300 2005](1) its velocity will decrease (2) its velocity will increase(3) it will turn towards right of direction of motion (4) it will turn towards left of direction of motion.

12. A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to amagnetic field B. The time taken by the particle to complete one revolution is : [AIEEE 4/300 2005]

(1) Bmq2

(2) B

Bq2 2

(3) mqB2

(4) qBm2

13. A magnetic needle is kept in a non-uniform magnetic field. It experiences : [AIEEE 4/300 2005](1) a torque but not a force (2) neither a force nor a torque(3) a force and a torque (4) a force but not a torque

14. In a region, steady and uniform electric and magnetic fields are present. These two fields are parallel to each other.A charged particle is released from rest in this region. The path of the particle will be a : [AIEEE 1.5/180 2006](1) circle (2) helix (3) straight line (4) ellipse

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 35: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 34

15. Needles N1, N

2 and N

3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respec-

tively. A magnet when brought close to them will : [AIEEE 1.5/180 2006](1) attract all three of them(2) attract N

1 and N

2 strongly but repel N

3

(3) attract N1 strongly, N

2 weakly and repel N

3 weakly

(4) attract N1 strongly, but repel N

2 and N

3 weakly

16. A long solenoid has 200 turns per cm and carries a current i. The magnetic field at its centre is 6.28×10�2

Weber/m2. Another long solenoid has 100 turns per cm and it carries a current i/3. The value of the magneticfield at its centre is : [AIEEE 4.5/180 2006](1) 1.05 × 10�4 Weber/m2 (2) 1.05 × 10�2 Weber/m2

(3) 1.05 × 10�5 Weber/m2 (4) 1.05 × 1010�3 Weber/m2

17. A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its

crosssection. The ratio of the magnetic field at 2a

and 2a from axis is : [AIEEE 3/120 2007]

(1) 1/4 (2) 4 (3) 1 (4) 1/2

18. A current I flows along the length of an infinitely long, straight, thin walled pipe. Then : [AIEEE 3/120 2007](1) the magnetic field is zero only on the axis of the pipe(2) the magnetic field is different at different points inside the pipe(3) the magnetic field at any point inside the pipe is zero(4) the magnetic field at all points inside the pipe is the same, but not zero

19. A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E

and

B

with a velocity v perpendicular to both E

and B

, and comes out without any change in magnitude or

direction of v

. Then : [AIEEE 3/120 2007]

(1) 2B/BEv

(2) 2B/EEv

(3) 2E/EEv

(4) 2E/EBv

20. A charged particle moves through a magnetic field perpendicular to its direction. Then : [AIEEE 3/120 2007](1) the momentum changes but the kinetic energy is constant(2) both momentum and kinetic energy of the particle are not constant(3) both, momentum and kinetic energy of the particle are constant(4) kinetic energy changes but the momentum is constant

21. Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries anelectric current

and COD carries a current

. The magnetic field on a point lying at a distance d from O, in

a direction perpendicular to the plane of the wires AOB and COD, will be given by : [AIEEE 3/120 2007]

(1) 2/1

210

d2

(2) 2/12

221

0

d2

(3)

d20

(1 + 2) (4) 2

221

0

d2

22. Relative permittivity and permeability of a material are r and

r, respectively. Which of the following values of

these quantities are allowed for a diamagnetic material ? [AIEEE 3/105 2008](1)

r = 1.5 ,

r = 0.5 (2)

r = 0.5 ,

r = 0.5 (3)

r = 1.5 ,

r = 1.5 (4)

r = 0.5 ,

r = 1.5

23. A horizontal overhead powerline is at a height of 4 m from the ground and carries a current of 100 A from eastto west. The magnetic field directly below it on the ground is (

0 = 4 × 10�7 T mA�1):

[AIEEE 3/ 105 2008](1) 5 × 10�6 T northward (2) 5 × 10�6 T southward(3) 2.5 × 10�7 T northward (4) 2.5 × 10�7 T southward

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 36: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 35

Comprehension : [AIEEE 2009 ]

Direction : Question numbers 24 and 25 are based on the following paragraph :

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b)and DA (radius = a) of the loop are joined by two straight wires AB and. CD. A steady current is flowing inthe loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current

flowing out of the plane of the paper is kept at the origin.

24. The magnitude of the magnetic field due to the loop ABCD at the origin (O) is: [AIEEE 4/144 2009]

(1) ab24

)ab(0 (2)

abab

40

(3)

)ba(

3)ab(2

40

(4) zero

25. Due to the presence of the current 1 at the origin : [AIEEE 4/144 2009]

(1) The forces on AD and BC are zero.

(2) The magnitude of the net force on the loop is given by

)ba(

3)ab(2

410

(3) the magnitude of the net force on the loop is given by ab24

10 (b � a).

(4) the forces on AB and DC are zero.

26. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of the planeof the paper as shown. The variation of magnetic field B along the line XX´ is given by [AIEEE 4/144 2010]

(1) (2)

(3) (4)

27. A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. Themagnitude of the the magnetic induction along its axis is : [AIEEE 4/144 2011]

(A) R

µ

20

I(B)

R2

µ

20

I(C)

R2

µ0

I(D)

R4

µ0

I

28. Proton, Deutron and alpha particle of the same kinetic energy are moving in circular trajectories in a constantmagnetic field. The radii of proton deuteron and alpha particle are respectively r

p, r

d, and r

. Which of the

following relation is correct ? [AIEEE 4/120 2012](1) r

= r

d >

r

p(2) r

= r

p >

r

d(3) r

= r

p <

r

d(4) r

> r

d >

r

p

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 37: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 36

NCERT QUESTIONS

1. Answer the following questions ;(a) What happens if a bar magnet is cut into two pieces (i) transverse to its length (ii) along its length?(b) What happens if an iron bar magnet is melted? Does it retain its magnetism?(c) A magnetized needle in a uniform magnetic field experiences a torque but no net force. An iron nailnear a bar magnet, however, experiences a force of attraction in addition a toroid. Why?(d) Must every magnetic field configuration have a north pole and a south pole? What about the fielddue to a toroid?(e) Can you think of a magnetic field configuration with three poles?(f) Two identical looking iron bars A and B are given , one of which is definitely known to be magnetized.(We do not know which one.) How would one ascertain whether or not both ate magnetized? If only oneis magnetized, how does one ascertain which one? [Use nothing else but the two bars A and B]

2. Answer the following questions regarding earth�s magnetism :

(a) A vector needs three quantities for its specification. Name the three independent quantitiesconventionally used to specify the earth�s magnetic field.

(b) The angle of dip at a location in southern India is about 18º. Would you expect a greater of smaller

dip angle in Britain?(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go intothe ground or come out of the ground?

3. A short bar magnet placed with its axis at 30º with a uniform external magnetic field of 0.25 T experiences

a torque of magnitude equal to 4.5 x 10�2 J. What is the magnitude of magnetic moment of the magnet?

4. A short bar magnet of magnetic moment m = 0.32 JT �1 is placed in a uniform external magnetic field of0.15 T. If the bar is free to rotate in the plane of the field, which orientations would correspond to its (i)stable and (ii) unstable equilibrium? What is the potential energy of the magnet in each case?

5. A closely wound solenoid of 800 turns and area of cross-section 2.5 x 10�4 m2 carries a current of 3.0A Explain the sense in which the solenoid acts like a bar magnet. What is its associated magneticmoment ?

6. A bar magnet of magnetic moment 1.5 JT�1 lies aligned with the direction of a uniform magnetic field of0.22 T.(a) What is the amount of work required by an external torque to turn the magnet so as to align itsmagnetic moment, (i) normal to the field direction (ii) opposite to the field direction?(b) What is the torque on the magnet in cases (i) and (ii)?

7. A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 10�4 m2 , carrying a current of4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.(a) What is the magnetic moment associated with the solenoid?(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x10�2 T isset up at an angle of 30º with the axis of the solenoid ?

8. At a certain location in Africa, a compass points 12º west of the geographic north. the north tip of the

magnetic needle of a dip circle placed in the plane of magnetic meridian points 60º above the horizontal.

The horizontal component of the earth�s field is measured to be 0.16 G. Specify the direction and

magnitude of the earth�s field at the location.

9. A monoenergetic ( 18 keV ) electron beam initially in the horizontal direction is subject to a horizontalmagnetic field of 0.40 G. normal to the initial direction. Estimate the up or down deflection of the beamover a distance of 30 cm,( m

e= 9.11 x 10�31 kg, e = 1.60 x 10�19 C ).

[Note : Data in this exercise are so chosen that the answer will give you an idea of the effect of earth�smagnetic field on the motion of the electron beam from the electron gun to the screen in a TV set]

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 38: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 37

Exercise # 1PART-I

A-1.* (ABC) A-2. (B) A-3. (A) B-1. (A) B-2. (B) B-3. (C) C-1. (A)

C-2. (B) C-3. (D) C-4. (C) C-5. (A) C-6. (D) C-7. (B) D-1. (B)

D-2. (D) D-3. (C) D-4. (B) D-5. (C) D-6. (B) D-7.* (AC) D 8.* (CD)

E-1. (D) E-2.* (AD) E-3.* (BD) E 4*. (BD) E-5.* (AB) F-1. (C) F-2. (C)

F-3. (B) F-4. (B) F-5. (D) G-1. (B) G-2. (A) H-1. (C) H-2. (A)

-1. (C)PART-II

1. (C) 2. (A) 3. (A) 4. (A) � p,q, r ; (B) � p, q, r, s ; (C) � r ; (D) � p, q, r, s

5. (A) � p, q ; (B) � p, r ; (C) � p ; (D) � p, q, s 6. (i) � R ; (ii) � Q, V ; (iii) � V ; (iv) � U

Exercise # 2PART-I

1. (C) 2. (C) 3. (A) 4. (A) 5. (D) 6. (C) 7. (A)

8. (D) 9. (B) 10. (A) 11. (D) 12. (C) 13. (C) 14.* (CD)

15.* (AD) 16.* (BCD) 17.* (BCD) 18.* (ABD) 19.* (ABC)

PART-II

1..13

2 10 -4 wb/m2

2. (i) 0, (ii) 10�13 k� (iii) �10�13 j� (iv) 274

× 10�13 (�2 j� + k� ) (v) yes, no (vi) yes, no.

3. (i) 20

R4

qv

, inwards (ii)

)Rx(4

qv22

0

, No

4. 1 × 10-4 wb/m2, towards the reader 5. 4 × 10-5 wb/m2

6. (i) 32 × 10�5 tesla (ii) 2 × 10�5 T 7. 0

8.d4

i0

, 2 d4

i0

ðì

, 2

19.

x2

i0

10.a

i22 0

11. 0

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 39: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 38

12. (i) (a) 4 × 10�4 (b) zero (ii) �410 x 22 T 13. (a)

1313

81105 5 T (b) zero

14.

21

0

R1

R1

4 15. B = 1222R4i 20

16. (a) B =

ba2

40

; (b) B =

b2

a23

40

(c) B = ( � + tan ) 0 /2R = 28T..

17. (a) )k�2i�(R4

B 0

(b)

k�11i�

R4B 0

(c) k�j�R4

B 0

18. B =2

ni0 19. B = 20

b2

i

r 20. (a) zero (b)

r5

i2 0

21. 2500 turns/m 22. 1 V 23. 12 cm 24. D � electron , B � -particle

25. M�3 L�2 T4 Q4

26. (a) 4 A, (b)(i) current directed into the plane of paper, 1 m from R on RQ (away from Q)(ii) current directed out of from paper, 1 m from R on RQ (between R and Q)

27. 30º 28. (�75 i� + 100 j� ) m/s 29. 3.0 40. 20 cm,13 s10

5

41. 2 cm 42.4

182× 10�4 T,

810x91

8

per sec 43. 1210x

5864

m/s2

44. (a) 1210x

918

m/s, No (b) 48 × 106 s

45. (a) qBmv

(b) (c) qB2m

(d) 2

3,

qBmv

qB2m3

46. (a) /2 (b) /4 (c) 47. 8 cm

48. 36 cm, 56 cm 49.

4 × 10 5 m/s, 2 × 10 5 m/s

50. 16 × 106 m/s, 2091

cm 51. Less than, same as

52. (a) evB (b) vB (c) vB

53. (a)ner

i2

(b)

nr

iB2

upwards in the figure (c)

ner

iB2

(d)

rneiB2

54. 50 m/s 55.45

× 10 5 C/kg 56. 5 × 10 3 N/C. 5 × 10�2 T

57.m

xqE2 0 58. 6 × 10�2 N perpendicular to both the wire and the field

59. 1 × 10�2 N on each wire, on da and cb towards left and on dc and ab downward.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 40: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 39

60. 8 × 10�2 N 61. iB2 0 62. 2 N towards the inside of the circuit

63. 5 × 10�4 T, horizontal and to the wire 64.1

2210

r

rn

2

ii

= 40 n2 J/m

65. iB/2 66. � iRB j� , upward in the figure

67. 0.12 68. iB0

69. 2 cm from the 20 A current and 8 cm from the other 70. 1.5 × 10�3 N/m, downward zero

71. 40 cm . 75 × 10�3 J

73. (a) 2aiB, perpendicular to the plane of the figure going into it. (b) 0Bai2

.21

T . (a) 4 × 10�2 N-m (b) 60º

. (a) zero (b) 2 × 10�2 N-m parallel to the side. . × 10�2 N - m

78. (a)4BiL2

(b) 18

BiL2

79. M = 4 × 10�24 A-m2, B = 32 / 5 = 6.4Wb/m2 80.21

QR2

81. V =

278kA,

H = kA

3139

82. 313 × 10 -5 T, 30º E of N

83. (a) 1.0 A , (b) 2.0 V perpendicular to the magnetic meridian

84. 6× 10�4 T, 4105

36

TT 85.

a

i22 0

86. d =

r387. B =

2

j )(0

88.13

× 10�5 T; 2× 10�6 TT 89. (a) V

max =

m2qBd

(b)maxV12d

(c) 3 d

90. 21i

mg

91. F =

a1n)ii(

2 00

in the direction of i0. x =

na

1 , where x is the perpendicular distance from

the wire i0. It will try to become antiparallel to i0.

92. i =mg

BN2 = 2.5 A

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 41: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 40

Exercise # 3PART-I

1. (A) 2. (D) 3. RP : R = 1 : 2 4. (B)

5. (i) K = NAB (ii)

0NAB2(iii)

C

NABQ

6.* (AC) 7. (A)

8. (A) � q, r ; (B) � p ; (C) � q, r ; (D) � q,s 9. (ACD)

10. (A) � p, r, s; (B) � r, s; (C) � p, q, t ; (D) � r, s 11. 7

12. (C) 13.* (BD) 14. (A) 15.* (CD) 16. 5 17. (B)

18. (D) 19. (AC) 20. (AD)

PART-II

1. (1) 2. (4) 3. (2) 4. (2) 5. (1) 6. (3) 7. (2)

8. (3) 9. (1) 10. (3) 11. (1) 12. (4) 13. (3) 14. (3)

15. (3) 16. (2) 17. (3) 18. (3) 19. (1) 20. (1) 21. (2)

22. (1) 23. (2) 24. (1) 25. (1) 26. (1) 27. (1) 28. (2)

Exercise # 4

1. (a) In either case, one gets two magnets, each with a north and south pole,(b) Molten iron is above the Curie temperature (770 ºC) and is, therefore, not ferromagnetic. An iron bar

magnet when melted does not does not retain its magnetism.(c) No force if the field is uniform. The iron nail experiences a non-uniform magnetic field due to the barmagnet. The induced magnetic moment in the nail, experiences both force and torque. The net force isattractive because the induced (say) south pole in the nail is closer to the north pole of the magnet than theinduced north pole.(d) Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not sofor a toroid of even for a straight infinite conductor.(e) Depends on what one means by three poles. Poles must always occur in pairs. But one can think to twobar magnets with (say) their north ends glued together as providing a three-pole field configuration.(f) Try to bring different ends of the magnets closer. A repulsive force in some situation establishes that bothare magnetized. If it is always attractive, then one of them is not magnetized. To see which one, pick up one,say , A and lower one of the middle of B is magnetized. If you do not notice any change from the end to themiddle of B, then A is magnetized.

2. (a) Magnetic declination, angle of dip, horizontal component of earth�s magnetic field.

(b) Greater in Britain (it is about 70º) , because Britain is closer to the magnetic north pole.

(c) Field lines of B due to the earth�s magnetism would seen to come out of the ground.

3. 0.36 JT�1

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303

Page 42: Contentsfile.etoosindia.com/sites/default/files/studymaterials/Electro Magnetic Field.pdfContents ELECTRO MAGNETIC FIELD (EMF) Syllabus Biot-Savarts law and Ampere™ ™s law ; Magnetic

EMF (JEE ADVANCED) # 41

4. (a) m parallel to B ; U = �mB = �4.8 x 10�2 J ; stable(b) m anti-parallel to B ; U� = +mB = +4.8 x 10�2 J ; unstable

5. 0.60 JT�1 along the axis of the solenoid ; the direction determined by the sense of flow of current.

6. (a) (i) 0.33 J (ii) 0.66 J(b) (i) torque of magnitude 0.33 J in a direction that tends to align the magnetic moment vector B. (ii)Zero.

7. (a) 1.28 A m2 along the axis in the direction related to the sense of current via the right-handed screwrule.(b) force is zero in uniform field; torque= 0.048 Nm in a direction that tends to align the axis of thesolenoid ( i.e., its magnetic moment vector) along. B.

8. The earth�s field lines in a vertical 12º west of the geographic meridian making an angle of 60º (

upwards ) with the horizontal (magnetic south to magnetic north) direction. Magnitude = 0.32 G.

9. R =eB

meV

eB

energykineticxm2 e = 11.3 m

Up or down deflection R (1 �cos ) where sin = 0.3 / 11.3. We get deflection ~ 4 mm.

ETOOSINDIA.COMIndia's No.1 Online Coaching for JEE Main & Advanced

3rd Floor, H.No.50 Rajeev Gandhi Nagar, Kota, Rajasthan 324005 HelpDesk : Tel. 092142 33303