Moving Charges and Magnetism - 1 File Download

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Level - II

SECTION - A

Objective Type Questions

(Magnetic Force, Motion of Charged Particle in a Magnetic Field)

1. A charged particle enters a uniform magnetic field perpendicular to it. The magnetic field

(1) Increases the speed of the particle

(2) Decreases the kinetic energy of the particle

(3) Changes the direction of motion of the particle

(4) Both (1) & (3)

Sol. Answer (3)

Magnetic force F v��

No work is done by magnetic field so speed and kinetic energy cannot be changed by magnetic field but

it can deflect the particle

2. A proton and an alpha particle enter the same magnetic field which is perpendicular to their velocity. If they have

same kinetic energy then ratio of radii of their circular path is

(1) 1 : 1

(2) 1 : 2

(3) 2 : 1

(4) 1 : 4

Sol. Answer (1)

2mkr

qB here k is same, so

mr

q

so 1 2

1 : 14 1

p p

p

r m q

r m q

Solutions

Chapter 4

Moving Charges and Magnetism

120 Moving Charges and Magnetism Solutions of Assignment (Level-II)


3. A particle of charge –q and mass m enters a uniform magnetic field B�

at A with speed v1 at an angle and

leaves the field at C with speed v2 at an angle as shown. Then

× × ×

× × ×

× × ×

× × ×

× × ×

× × ×

v1

A

C

v2

(1) =

(2) v1 = v

2

(3) Particle remains in the field for time t = qB

m )(2

(4) All of these

Sol. Answer (4)

=

1 2

mF vv v ∵

2 –

Tm

qB

4. A particle of charge per unit mass is released from origin with a velocity 0ˆv v i in a uniform magnetic field

0ˆB B k . If the particle passes through (0, y, 0) then y is equal to

(1)0

0

2v

B

(2)0

0

v

B (3)0

0

2v

B (4)0

0

v

B

Sol. Answer (3)

q

m

2R

(0, , 0)y

x0

0

22 2

mv vy R

qB B

(Motion in Combined Electric and Magnetic Fields)

5. A proton moving with a constant velocity, passes through a region of space without change in its velocity. If E

& B represent the electric and magnetic fields respectively, this region may have

(1) E = 0, B 0 (2) E 0, B = 0

(3) E & B both parallel (4) E & B inclined at 45° angle

Sol. Answer (1)

121Solutions of Assignment (Level-II) Moving Charges and Magnetism


6. A proton is accelerating in a cyclotron where the applied magnetic field is 2 T. If the potential gap is effectively

100 kV then how much revolutions the proton has to make between the “dees” to acquire a kinetic energy of 20

MeV?

(1) 100 (2) 150 (3) 200 (4) 300

Sol. Answer (1)

Energy increased in each revolution = 2 100 103 eV

= 2 105 eV

Now for energy E = 2 107 eV

Number of revolution =

7

5

2 10 eV100

2 10 eV

(Magnetic Field due to a Current Element : Biot-Savart Law)

7. A current i ampere flows in a circular arc of wire which subtends an angle 2

3 radian at its center, whose radius is

R. The magnetic field B at its center is

(1)R

i0(2)

R

i

2

3 0(3)

R

i

4

3 0(4)

R

i

8

3 0

Sol. Answer (4)

0 0

3 3

22 82

i iB

R R

8. At what distance on the axis, from the centre of a circular current carrying coil of radius r, the magnetic field

becomes 1/8th of the magnetic field at centre?

(1) r2 (2) r2/3

2 (3) r3 (4) r23

Sol. Answer (3)

2

0 0

3 22 2

1

8 2 2

i lr

r r x

3 2 32 2 8rr x

2 332 2

8rr x

2 2 24r x r

3r2 = x2

3x r

9. Magnetic field at P due to given structure is

P

I

I

R

R

R

I

(1)0

4 2

I

R

(2)0 6

4 5

I

R

(3)

0 5

4 6

I

R

(4)0 2

4

I

R



Sol. Answer (3)

0 0 0

–4 4 43 2

p

I I IB

R R R

��

0 051 1

1 –4 6 43 2

p

I IB

R R

��

10. A straight wire of finite length carrying current I subtends an angle of 60° at point P as shown. The magnetic field

at P is

60°

x x

I

P

(1)x

I

32

0

(2)x

I

2

0(3)

x

I

2

3 0(4)

x

I

33

0

Sol. Answer (1)

0sin30 sin30

4 cos30

IB

x

0 1 1

2 234

2

IB

x

x 30°

30°

0

2 3

i

x

11. Magnetic field at the centre O due to the given structure is

I

R

OR

I

I

(1)0 3 1

4 2

I

R

� (2)

0 13

2

I

R

(3)

0 3 1

4 2

I

R

(4)

0 23

4

I

R

�

Sol. Answer (3)

B = Bdue to circular arc

+ Bdue to straight wires

0 0

3

2.2 42

i i

R R

0 3 1

4 2

iB

R



12. A current i flows in a thin wire in the shape of a regular polygon with n sides. The magnetic induction at the centre

of the polygon when n is (R is the radius of its Circumcircle)

(1)6

tan2

0

R

in(2)

nR

in

tan

2

0(3)

R

i

2

0 (4) Zero

Sol. Answer (3)

For regular polygon having n sides where n will be almost a circle

So 0

2

iB

R

13. Two long straight wires are placed along x-axis and y-axis. They carry current I1 and I

2 respectively. The

equation of locus of zero magnetic induction in the magnetic field produced by them is

(1) y = x (2)2

1

Iy x

I

(3)1

2

Iy x

I

(4) y = (I1I2)x

Sol. Answer (3)

x

y

( = 0)B

A

l2

l1

On a general point A magnetic field will be zero when 0 2 0 1

2 2

l l

x y

1

2

ly x

l

14. Surface charge density on a ring of radius a and width d is as shown in the figure. It rotates with frequency

f about its own axis. Assume that the charge is only on outer surface. The magnetic field induction at centre

is(Assume that d << a)

d

(1) 0fd (2) df 0 (3) df 02 (4) df

0

2

2

Sol. Answer (1)

Surface charge density =

Total charge on the ring (q) = 2 da

2T

qi dfa

0 0

0

2

2 2

l adfB df

a a

��



15. A current of i ampere is flowing in an equilateral triangle of side a. The magnetic induction at the centroidwill be

(1)a

i

33

0(2)

a

i

2

3 0(3)

a

i

3

25 0(4)

a

i

2

9 0

Sol. Answer (4)

O is centroid and using the OAD distance OD = 2 3

a

60° 60°

A

C

BD2a

O

By all the three sides AB, BC and CA, direction of magnetic

field produced will be same and inward to the plane of paper

So 0

total

sin60 sin603

42 3

i

aB

= 0

9

2

i

a

(Magnetic Field on the Axis of a Circular Current Loop)

16. The magnetic field intensity at the point O of a loop with current i, whose shape is illustrated below is

O

a

i

b

b

(1)

ba

i 2

2

3

4

0

(2)

ba

i 2

42

0(3)

ba

i 11

2

0(4)

ba

i 11

4

0

Sol. Answer (1)

B

��

due to square part :-

B

��

due to side OA and OC will be zero at point O

B

��

due to side AB and BC will be equal so

0

12 sin45 0

4

iB

b

��

0

2 2

i

b

b

b

OD

C

BA

45°

45°i

B

��

due to circular part

0 0

2

33

22 8

2

i iB

a a

��

net 1 2 0

3 1

8 2 2B B B i

a b

��

0 3 2

4 2

i

a b

17. A square frame of side l carries a current i. The magnetic field at its centre is B. The same current is passedthrough a circular coil having the same perimeter as the square. The field at the centre of the circular coil

is B. The ratio of B

B

is

(1)2

28

(2)

3

28

(3)

28

(4)2

24



Sol. Answer (1)

2

lOD

0sin45 sin45

44

2

i

B l

45° 45°

l

D C

BA

O

02 2

iB

l

For circle

2 4r l r2l

r

So 0 0'

222

i iB

lr

0'

4

iB

l

So 2

8 2

'

B

B

(Ampere’s Circuital Law, The Solenoid and the Toroid, Force between two Parallel Currents, the Ampere,Torque on Current Loop, Magnetic Dipole, The Moving Coil Galvanometer)

18. If 1 2 3, andB B B

� � �

are the magnetic field due to I1, I

2 and I

3, then in Ampere’s circuital law 0

· ,B dl I B � � �

� is

×

I1

I2

I3

(1)1 2–B B B

� � �

(2)1 2 3

B B B B � � � �

(3)1 2 3–B B B B

� � � �

(4)3

B B� �

Sol. Answer (2)

In ampere circuital law, on amperian loop B is due to all the current elements either inside or outside to theamperian loop

1 2 3B B B B � � � �

19. A charge Q moves parallel to a very long straight wire carrying a current I as shown. The force on the charge is

I

+ Q P O

Y

X

v

(1) Opposite to OX (2) Along OX (3) Opposite to OY (4) Along OY



Sol. Answer (1)

F q V B ��

Using right hand thumb rule F will be opposite to OX.

l O

Y

XY

v

F

20. A uniform conducting wire ABC has a mass of 10 g. A current of 2 A flows through it. The wire is kept in a

uniform magnetic field B = 2 T. The acceleration of the wire will be

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

4 cm

5 cm

A

B

C

(1) Zero (2) 12 ms–2 (3) 1.2 × 10–3 ms–2 (4) 0.6 × 10–3 ms–2

Sol. Answer (2)

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

4 cm

5 cm

A

B

C3 cm

Force on wire ABC will be same as force on wire AC,

using F i l B � ��

32 sin902

100F

–212 10 N

–2

2

–2

12 1012 m/s

10

Fa

m

21. Figure shows a conducting loop ADCA carrying current i and placed in a region of uniform magnetic field B0.

The part ADC forms a semicircle of radius R. The magnitude of force on the semicircle part of the loop is equal

to

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

A

C

DB0

(1) RiB0

(2) Zero (3) 2RiB0

(4) 2iRB0

Sol. Answer (4)

The force on the semicircle part ADC, will be same as force on wire CA and force on wire

CA = i (2R)(B0) (using F = ilB)

= 2iRB0



22. A wire carrying a current i is placed in a magnetic field in the form of the curve sin 0 2x

y a x LL

.

Force acting on the wire is

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

O

2L

(1)iBL

(2) iBL (3) 2iBL (4) Zero

Sol. Answer (3)

2Ll �

Now, using F ilB

2i BL

23. The magnetic field existing in a region is given by 0ˆ1

xB B k

l

�

. A square loop of edge l and carrying a

current i, is placed with its edge parallel to the x-y axes. Find the magnitude of the net magnetic forceexperienced by the loop

(1) 0

1

2iB l (2) Zero (3) iB

0l (4) 2iB

0l

Sol. Answer (3)

0

ˆ1x

B B kl

��

at x = 0, 1 0

ˆB B k

at x = l, 2 0

ˆ2B B kF

1F

2

Fnet

= 2 12 1–– B BF F il

0 02 –il B B

0ilB

24. Two protons A and B move parallel to the x-axis in opposite directions with equal speeds v. At the instant shown,the ratio of magnetic force and electric force acting on the proton A is (c = speed of light in vacuum)

d

ve

A

B

e

v

x

y

(1)c

v

(2)2

2

c

v

(3)c

vd2

(4)c

v2



Sol. Answer (2)

25. If the planes of two identical concentric coils are perpendicular and the magnetic moment of each coil is M,

then the resultant magnetic moment of the two coils will be

(1) M (2) 2 M (3) 3 M (4) M2

Sol. Answer (4)

2 2

netM M M

2M

26. In a hydrogen atom, an electron of mass m and charge e revolves in an orbit of radius r making n revolutions per

second. If the mass of hydrogen nucleus is M, the magnetic moment associated with the orbital motion of

electron is

(1)mM

mner

2

(2) ner2 (3)m

ner2

(4)M

mner2

Sol. Answer (2)

Magnetic moment = NiA

qi en

T

2A r

and N = 1

Magnetic moment 21 enm r

27. The phosphor bronze strip is used in a moving coil galvanometer because

(1) It is torsional constant is small (2) It is easily available

(3) It is paramagnetic (4) It is diamagnetic

Sol. Answer (1)

28. A uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the

centre due to ABC portion of the wire will be (length of ABC = l1, length of ADC = l

2)

D

C

OR

B

Ai

(1) 2

21

210

)(2 ll

lil

R

(2))(2 21

2

2

0

ll

il

R

(3)21

210 )(

2 ll

lli

R

(4) Zero

Sol. Answer (1)

Let current in part ABC is i1

and in part ADC is i2



2

1 2

ili

l l

(As ABC and ADC part are in parallel connection)

and subtended by ABC at centre O will be 11 2

2l

l l

so using 0

2 2

iB

a

120

1 2 1 2

2

2 2

lilB

l l l lR

SECTION - B

Previous Years Questions

1. A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the centre of the solenoid, a coil of 100

turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid

reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 102 , the total charge

flowing through the coil during this time is [NEET-2017]

(1) 32C (2) 16 C (3) 32 C (4) 16C

Sol. Answer (3)

dN

dt

N d

R R dt

Ndq d

R

( )NQ

R

totalQR

( )NBA

R

2

0ni r

R

Putting values

7 2

2

4 10 100 4 (0.01)

10

32 CQ

2. An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current

‘I’ along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire ‘B’ is given

by [NEET-2017]

90°

CB

A

d

d

(1)

2

0

2

I

d

(2)

2

02 I

d

(3)

2

02 I

d

(4)

2

0

2

I

d



Sol. Answer (4)

Force between BC and AB will be same in magnitude.

90°

C

B

A

F

d

F

d

2

0

2BC BA

IF F

d

2BC

F F

2

022

I

d

2

0

2

IF

d

3. A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre

of the loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of

n turns will be [NEET-(Phase-2)-2016]

(1) nB (2) n2B (3) 2nB (4) 2n2B

Sol. Answer (2)

0

2

IB

r

, when made n turns radius becomes r '

2 ' 2 'r

n r r r

n

Now, 2 20 0

'2 ' 2

nI IB n n B

r r

4. An electron is moving in a circular path under the influence of a transverse magnetic field of

3.57 × 10–2 T. If the value of e/m is 1.76 × 1011 C/kg, the frequency of revolution of the electron is

[NEET-(Phase-2)-2016]

(1) 1 GHz (2) 100 MHz (3) 62.8 MHz (4) 6.28 MHz

Sol. Answer (1)

f = 11 2

1.76 10 3.57 10

2 2 3.14

qB

m

= 109 Hz = 1 GHz

5. A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform

magnetic field of strength 0.2 Weber/m2. The coil carries a current of 2 A. If the plane of the coil is inclined

at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will

be [Re-AIPMT-2015]

(1) 0.12 Nm (2) 0.15 Nm (3) 0.20 Nm (4) 0.24 Nm

Sol. Answer (3)

Torque on coil, = nIABsin60°

350 2 (0.12 0.1) 0.2

2 = 0.20 Nm



6. A proton and an alpha particle both enter a region of uniform magnetic field B, moving at right angles to the

field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton

is 1 MeV, the energy acquired by the alpha particle will be [Re-AIPMT-2015]

(1) 1 MeV (2) 4 MeV (3) 0.5 MeV (4) 1.5 MeV

Sol. Answer (1)

r = 2mE

qB

pp p pmr q E

r q Em

1p

E

E

1p

E

E

7. A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long

and parallel to X-axis while semicircular portion of radius R is lying in Y-Z plane. Magnetic field at point O is

[AIPMT-2015]

Z

I

YO

I

X

I

R

(1) 0 ˆ ˆ24

IB i k

R

�

(2) 0 ˆ ˆ24

IB i k

R

�

(3) 0 ˆ ˆ24

IB i k

R

�

(4) 0 ˆ ˆ24

IB i k

R

�

Sol. Answer (4)

0 0 0ˆ ˆ ˆ ˆ2 24 4 4

� I I IB k i i k

R R R

8. An electron moving in a circular orbit of radius r makes n rotations per second. The magnetic field produced

at the centre has magnitude [AIPMT-2015]

(1)0

2

ne

r

(2)

0

2

ne

r

(3) Zero (4)

20n e

r

Sol. Answer (1)

B = 0

2

i

r

i = e × n

B = 0

2

e n

r



9. Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other

such that O is their common point for the two. The wires carry I1 and I

2 currents, respectively. Point 'P' is lying

at distance d from O along a direction perpendicular to the plane containing the wires. The magnetic field at the

point P will be [AIPMT-2014]

(1)0 1

22

I

d I

(2) 0

1 22

I Id

(3) 2 20

1 22

I Id

(4) 1/22 20

1 22

I Id

Sol. Answer (4)

Since the wires are perpendicular, their magnetic fields also are perpendicular. So the resultant field will be

pythagoras of both the fields.

10. A current loop in a magnetic field [NEET-2013]

(1) Can be in equilibrium in one orientation

(2) Can be in equilibrium in two orientations, both the equilibrium states are unstable

(3) Can be in equilibrium in two orientations, one stable while the other is unstable

(4) Experiences a torque whether the field is uniform or non-uniform in all orientations

Sol. Answer (3)

11. When a proton is released from rest in a room, it starts with an initial acceleration a0 towards west. When it is

projected towards north with a speed v0 it moves with an initial acceleration 3a

0 toward west. The electric and

magnetic fields in the room are: [NEET-2013]

(1)0 0

0

2 west, down

ma ma

e ev(2)

0 0

0

3 east, up

ma ma

e ev

(3)0 0

0

3 east, down

ma ma

e ev(4)

0 0

0

2 west, up

ma ma

e ev

Sol. Answer (1)

Case-I Case-II

a0

+e +e

v0

3a0

Now, 0

westeE

am

��

��

0 west

maE

e

��

��

Now, 0 02

BaF m ev B

0

0

2

maB

ev

12. Two similar coils of radius R are lying concentrically with their planes at right angles to each other. The currents

flowing in them are I and 2I, respectively. The resultant magnetic field induction at the centre will be

[AIPMT (Prelims)-2012]

(1)0

2

I

R

(2)

0I

R

(3) 0

5

2

I

R

(4)

03

2

I

R

Sol. Answer (3)



13. An alternating electric field, of frequency , is applied across the dees (radius = R) of a cyclotron that is being

used to accelerate protons (mass = m). The operating magnetic field (B) used in the cyclotron and the kinetic

energy (K) of the proton beam, produced by it, are given by [AIPMT (Prelims)-2012]

(1)2 m

Be

and K = 2m22R2 (2)m

Be

and K = m2R2

(3)m

Be

and K = 2m22R2 (4)2 m

Be

and K = m2R2

Sol. Answer (1)

mVR

qB here

eBRq e v

m

frequency 1 1

2 mT

eB

2 m

Be

2 2 2 2

21 1

2 2 2

e B ReBRk mv m

mm

22 2

2 2 222

2

e R mk m R

m e

14. A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What

should be the energy of an -particle to describe a circle of same radius in the same field?

[AIPMT (Mains)-2012]

(1) 2 MeV (2) 1 MeV (3) 0.5 MeV (4) 4 MeV

Sol. Answer (2)

15. A current carrying closed loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic

field acting along AB. If the magnetic force on the arm BC is F�

, the force on the arm AC is


B

A

C

(1) 2F�

(2) – 2F�

(3) –F�

(4) F�

Sol. Answer (3)

Net force on the loop = 0

Force on wire AB is zero because it is along B field. Hence Foce on AC = –(Force on BC) = –F�

B

A

C

i



16. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an

electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron


(1) Will turn towards left of direction of motion

(2) Will turn towards right of direction of motion

(3) Speed will decrease

(4) Speed will increase

Sol. Answer (3)

17. Charge q is uniformly spread on a thin ring of radius R. The ring rotates about its axis with a uniform frequency

f Hz. The magnitude of magnetic induction at the center of the ring is [AIPMT (Mains)-2011]

(1)0

2

q

fR

(2)

0

2

qf

R

(3)

0

2

qf

R

(4)

0

2

q

fR

Sol. Answer (3)

B = 0

2

qf

R

18. A square loop, carrying a steady current I, is placed in a horizontal plane near a long straight conductor carrying

a steady current I1 at a distance d from the conductor as shown in figure.The loop will experience

[AIPMT (Mains)-2011]

d

I1

I

I

(1) A net torque acting downward normal to the horizontal plane

(2) A net attractive force towards the conductor

(3) A net repulsive force away from the conductor

(4) A net torque acting upward perpendicular to the horizontal plane

Sol. Answer (2)

F1

F2

l1

net 1 2–F F F =

1 2F F

19. A square current carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the

force on one arm of the loop is F, the net force on the remaining three arms of the loop is:


(1) 3F�

(2) F�

(3) 3F�

(4) F�

Sol. Answer (2)



20. A beam of cathode rays is subjected to crossed electric (E) and magnetic fields (B). The fields are adjusted

such that the beam is not deflected. The specific charge of the cathode rays is given by (Where V is the potential

difference between cathode and anode) [AIPMT (Prelims)-2010]

(1)

2

22

B

VE(2)

2

2

2VB

E(3)

2

2

2VE

B(4)

2

22

E

VB

Sol. Answer (4)

21. A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the

other in x-z plane. If the current in the loop is i. The resultant magnetic field due to the two semicircular parts

at their common centre is [AIPMT (Mains)-2010]

(1)0

2 2

i

R

(2) 0

2

i

R

(3)

0

4

i

R

(4)

0

2

i

R

Sol. Answer (1)

22. A particle having a mass of 10–2 kg carries a charge of 5 × 10–8C. The particle is given an initial horizontal

velocity of 105 ms–1 in the presence of electric field E�

and magnetic field B�

. To keep the particle moving in a

horizontal direction, it is necessary that

(a) B�

should be perpendicular to the direction of velocity and E�

should be along the direction of velocity.

(b) Both B�

and E�

should be along the direction of velocity.

(c) Both B�

and E�

are mutually perpendicular and perpendicular to the direction of velocity.

(d) B�

should be along the direction of velocity and E�

should be perpendicular to the direction of velocity.

Which one of the following pairs of statements is possible? [AIPMT (Mains)-2010]

(1) (a) and (c) (2) (c) and (d) (3) (b) and (c) (4) (b) and (d)

Sol. Answer (3)

m

F q V B ��

eF qE

��

If m e

F F , particle will continuously move in horizontal direction

If B��

is in the direction of velocity Fm

= 0

23. A rectangular, a square, a circular and an elliptical loop, all in the (x-y) plane, are moving out of a uniform magnetic

field with a constant velocity ˆV vi�

. The magnetic field is directed along the negative z-axis direction. The

induced emf, during the passage of these loops, out of the field region, will not remain constant for


(1) The circular and the elliptical loops

(2) Only the elliptical loop

(3) Any of the four loops

(4) The rectangular, circular and elliptical loops

Sol. Answer (1)



24. The magnetic force acting on a charged particle of charge –2 C in a magnetic field of 2 T acting in

y-direction, when the particle velocity ˆ ˆ(2 3 )i j × 106 ms–1, is: [AIPMT (Prelims)-2009]

(1) 4 N in z-direction (2) 8 N in y-direction (3) 8 N in z-direction (4) 8 N in –z direction

Sol. Answer (4)

–2 q C ˆ2 B Tj

6ˆ ˆ 102 3V i j ��

–2 F C V B ��

–6 6

ˆ ˆ ˆ

–2 10 102 3 0

0 2 0

i j k

F

ˆ–8k

25. Under the influence of a uniform magnetic field, a charged particle moves with constant speed V in a circle of

radius R. The time period of rotation of the particle: [AIPMT (Prelims)-2009]

(1) Depends on R and not on V (2) Is independent of both V and R

(3) Depends on both V and R (4) Depends on V and not on R

Sol. Answer (2)

2 mT

qB

26. A particle of mass m, charge Q and kinetic energy T enters a transverse uniform magnetic field of induction B�

.

After 3 second the kinetic energy of the particle will be [AIPMT (Prelims)-2008]

(1) 4T (2) 3T (3) 2T (4) T

Sol. Answer (4)

B

��

does not change kinetic energy.

27. A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments

PS, SR and RQ are F1, F

2 and F

3 respectively and are in the plane of the paper and along the directions shown,

the force on the segment QP is [AIPMT (Prelims)-2008]

P

S R

F2

F1

F3

Q

(1) F3 – F

1 + F

2(2) F

3 – F

1 – F

2(3) 2

2

2

13 )( FFF (4) 2

2

2

13 )( FFF



Sol. Answer (3)

Net force on loop will be zero in uniform magnetic field

So, force on QP will balance other forces

2 2

3 1 2–

QPF FF F

28. A charged paritcle (charge q) is moving in a circle of radius R with uniform speed v. The associated magnetic

moment is given by [AIPMT (Prelims)-2007]

(1) q v R (2)2

qvR(3) q v R 2 (4)

2

2

qvR

Sol. Answer (2)

29. A beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields.If the electric

field is switched off, and the same magnetic field is maintained, the electrons move [AIPMT (Prelims)-2007]

(1) Along a straight line (2) In an elliptical orbit

(3) In a circular orbit (4) Along a parabolic path

Sol. Answer (3)

29. In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric

potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are

kept constant, the ratio charge on the ion

mass of the ion

will be proportional to [AIPMT (Prelims)-2007]

(1) R (2)1

R(3)

2

1

R(4) R2

Sol. Answer (3)

31. When a charged particle moving with velocity v�

is subjected to a magnetic field of induction B�

, the force on it

is non-zero. This implies that : [AIPMT (Prelims)-2006]

(1) Angle between v�

and B�

is necessarily 90°


and B�

can have any value other than 90°


and B�

can have any value other than zero and 180°


and B�

is either zero or 180°

Sol. Answer (3)

sinF q qvBv B � ��

0F , sin 0 , 0 , 180°

32. Two circular coils 1 and 2 are made from the same wire but the radius of the 1st coil is twice that of the 2nd coil.

What is the ratio of potential difference applied across them so that the magnetic field at their centres is the

same? [AIPMT (Prelims)-2006]

(1) 3 (2) 4 (3) 6 (4) 2

Sol. Answer (2)



33. A coil in the shape of an equilateral triangle of side l is suspended between the pole pieces of a permanent

magnet such that B�

is in plane of the coil. If due to a current i in the triangle a torque acts on it, the side l

of the triangle is : [AIPMT (Prelims)-2005]

(1)

1/22

3 Bi

(2)2

3 Bi

(3) 2

1/2

3Bi

(4)1

3 Bi

Sol. Answer (3)

34. A very long straight wire carries a current I. At the instant when a charge +Q at point P has velocity v�

, asshown, the force on the charge is : [AIPMT (Prelims)-2005]

I

P

Q

O X

Y

v

(1) Opposite to OX (2) Along OX (3) Opposite to OY (4) Along OY

Sol. Answer (4)

i

F q V B ��

��

using right hand thumb, F��

will be in the direction OY.

35. An electron moves in a circular orbit with a uniform speed v. It produces a magnetic field B at the centre of the

circle. The radius of the circle is proportional to [AIPMT (Prelims)-2005]

(1)B

v(2)

v

B(3)

v

B(4)

B

v

Sol. Answer (3)

0

34

q v rB

r

� �

0

24

qvB

r

vr

B

36. A beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength

20 Vm–1 and 0.5 T at right angles to the direction of motion of the electrons. What is the velocity of the

electrons?

(1) 8 ms–1 (2) 5.5 ms–1 (3) 20 ms–1 (4) 40 ms–1

Sol. Answer (4)

qvB qE

–120 40 ms

0.5

Ev

B



37. A 10 eV electron is circulating in a plane at right angles to a uniform field of magnetic induction

10–4 Wb/m2 (= 1.0 gauss), the orbital radius of electron is

(1) 11 cm (2) 18 cm (3) 12 cm (4) 16 cm

Sol. Answer (1)

2mkr

qB

38. A positively charged particle moving due East enters a region of uniform magnetic field directed vertically

upwards. This particle will

(1) Move in a circular path with a decreased speed (2) Move in a circular path with a uniform speed

(3) Get deflected in vertically upward direction (4) Move in circular path with an increased speed

Sol. Answer (2)

In uniform B��

, if charge enters perpendicular to the magnetic field. It will execute circular motion with uniform

speed.

39. An electron having mass m and kinetic energy E enters in uniform magnetic field B perpendicularly, then its

frequency will be

(1)`Bq

eE

(2)eB

m2(3)

m

eB

2 (4)eBE

m2

Sol. Answer (3)

f = m

eB

2

40. A positive charge particle with charge q is moving with speed v in a region of uniform magnetic field B at the

instant shown in figure. An external electric field is to be applied so that the charged particle follows a straight

line path. The magnitude and direction of electric field required are respectively

y

x

× × × ×

× × × ×

× × × ×

× × × ×

B

vq

(1) qvB, +y axis (2) qvB, –y axis (3) vB, –y axis (4)q

vB, –x axis

Sol. Answer (3)

ˆ

mF qvBj

for constant velocity

–B EF F��

ˆ–qvBj qE

��

ˆ–E vBj

��



41. The magnetic field of given length of wire for single turn coil at its centre is B then its value for two turns coil

for the same wire is

(1)4

B(2)

2

B(3) 4B (4) 2B

Sol. Answer (3)

0 0

22 2

2

i iB N B N

RR

04 4

2

iN B

R

42. The magnetic field dB due to a small current element dl at a distance r and element carrying current i is

(1)

r

rdlidB

�

20

4(2)

3

0

4 r

rdlidB

�

(3)

r

rdlidB

�

4

0 (4)

2

20

4 r

rdlidB

�

Sol. Answer (2)

0

34

i dl rdB

r

��

��

43. Two equal electric currents are flowing perpendicular to each other as shown in the figure. Lines AB and CD

are perpendicular to each other and symmetrically placed with respect to the currents. Where do we expect

the resultant magnetic field to be zero?

A

B

C

D

I

lO

(1) On CD (2) On AB

(3) On both OD and BO (4) On both AB and CD

Sol. Answer (2)

On line AB, as in this region magnetic field will be in opposite direction due to both the wires.

44. To convert a galvanometer into a voltmeter one should connect a

(1) High resistance in series with galvanometer

(2) Low resistance in series with galvanometer

(3) High resistance in parallel with galvanometer

(4) Low resistance in parallel with galvanometer

Sol. Answer (1)

45. A galvanometer having a coil resistance of 60 shows full scale deflection when a current of 1.0 A passes through

it. It can be converted into an ammeter to read currents upto 5.0 A by

(1) Putting in parallel a resistance of 15 (2) Putting in parallel a resistance of 240 (3) Putting in series a resistance of 15 (4) Putting in series a resistance of 240

Sol. Answer (1)

6015

5– 1 – 1

1g

RgRs

i

i



46. A straight wire of a diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter

carrying the same current. The strength of the magnetic field far away is

(1) One-quarter of the earlier value (2) No change

(3) Twice the earlier value (4) One-half of the earlier value

Sol. Answer (2)

For wire 1 d = 0.5 mm, r = 0.25 mm, l = 1A

For wire 2 d = 1 mm, r = 0.5 mm, l = 1A

0

2

iB

x

as x B = 0 (for both the cases)

47. A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10–4 m2 carries a current of

2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a

uniform magnetic field 5 × 10–2 tesla making an angle of 30º with the axis of the solenoid. The torque on the

solenoid will be

(1) 3 × 10–3 N.m (2) 1.5 × 10–3 N.m (3) 1.5 × 10–2 N.m (4) 3 × 10–2 N.m

Sol. Answer (3)

N = 2000 = 30°

–4 21.5 10 mA –2

5 10 TB l = 2A

M B ��

–2sin30 1.5 10 NmNlAB

48. In an ammeter 0.2% of main current passes through the galvanometer. If resistance of galvanometer is G, the

resistance of ammeter will be

(1)1

499G (2)

499

500G (3)

1

500G (4)

500

499G

Sol. Answer (3)

0.2 G = 100 RA R

A =

G

500

SECTION - C

Assertion-Reason Type Questions

1. A : Magnetic field lines are always perpendicular to the current producing it.

R : Magnetic field due to a straight wire varies in inverse square proportion with distance.

Sol. Answer (3)

2. A : Stationary charges do not experience a magnetic force.

R : Magnetic force is a central force.

Sol. Answer (3)

3. A : Net magnetic force experienced by a current carrying loop in a uniform magnetic field is always zero.

R : A current loop placed in a uniform magnetic field never experiences a torque.

Sol. Answer (3)

4. A : The trajectory of a charge when it is projected perpendicular to an electric field is a parabola.

R : A moving charge entering parallel to the magnetic field lines moves in a circular path.

Sol. Answer (3)



5. A : Like currents repel and unlike currents attract each other (in conductor).

R : Magnetic force acts in the direction of current.

Sol. Answer (4)

6. A : A magnetic dipole experiences maximum torque when it is placed normal to the magnetic field.

R : The minimum potential energy of magnetic dipole is zero.

Sol. Answer (3)

7. A : The relation between magnetic moment and angular momentum is true for every finite size body.

R : Ratio of magnetic dipole moment and angular momentum is just dependent on specific charge of the body.

Sol. Answer (1)

8. A : When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic

field is taken into account.

R : Magnetic field lines always form closed loops.

Sol. Answer (2)

9. A : Ampere's circuital law' is not independent of the Biot-Savart's law.

R : Ampere's Circuital law can be derived from the Biot-Savart law.

Sol. Answer (1)

10. A : The work done by magnetic field on a moving charge is zero.

R : The magnetic force acting on a moving charge has no component in the direction of velocity.

Sol. Answer (1)

11. A : In any magnetic field region the line integral dlB. along a closed loop is always zero.

R : The magnetic field B�

in the expression dlB. is due to the currents enclosed only by the loop.

Sol. Answer (4)

12. A : The magnetic field always accelerates a moving charge if the moving charge cuts the field lines.

R : When a moving charge cuts the magnetic field lines, the magnetic force on the charge is always non

zero.

Sol. Answer (1)

13. A : The magnetic moment of a current carrying planar loop does not depend on the shape of the loop.

R : The magnetic moment is a vector quantity.

Sol. Answer (2)

14. A : Magnetic field is produced by moving charge(s).

R : The magnetic field in the central region of a solenoid is uniform.

Sol. Answer (2)

15. A : In the middle to high latitudes on a dark night an aurora or the curtain of light hangs down from the sky.

This curtain is local, several hundred kilometer high, several thousand kilometer long but less than 1 km

thick.

R : Electrons and protons trapped in the helical terrestrial magnetic field collide with atoms and molecules

of air, causing that air to emit light.

Sol. Answer (1)

��

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