276

CHAPTER-15MAGNETICS

FILL IN THE BLANKS1. A neutron, a proton and an electron and alpha particle enter a region of constant magnetic field with equal velocities.

The magnetic field is along the inward normal to the plane of the paper. The tracks of the particles are labelled infigure. The electron follows track .... and the alpha particle follows track..... (1984; 2M)

A

BC

D

2. A wire of length L metres carrying a current i amperes is bent in the form of circle. The magnitude of its magneticmoment is .... in MKS units. (1987; 2M)

3. In a hydrogen atom, the electron moves in an orbit of radius 0.5 making 1016 revolutions per second. the magneticmoment associated with the orbital motion of the electron is ..... (1988; 2M)

4. The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown. Themagnitude of the magnetic induction at the centre C is ..... (1988; 2M)

R2

R Q PS

R1

5. A wire ABCDEF (with each side of length L) bent as shown in figure and carrying I is placed in a uniform magneticinduction B parallel to the positive y-direction. The force experienced by the wire is ..... in the ..... direction.(1990; 2M)

O

Bx

A

y

FE

IC

z

6. A metallic block carrying current I is subjected to a uniform magnetic

induction Bur

as shown in figure. The moving charges experience a force Fur

given by .... which results in the lowering of the potential of the face ....Assume the speed of the charge carriers to be V. (1996; 2M)

B YE

FX

GH

I

D

BA

C

Z

277

OBJECTIVE QUESTIONS

Only One option is correct :

1. A magnetic needle is kept in a non-uniform magneticfield. It experiences : (1981; 2M)(a) a force and a torque(b) a force but not a torque(c) a torque but not a force(d) neither a force nor a torque

2. A conducting circular loop of radius r carries a constant

current i. It is placed in a uniform magnetic field 0Bur

such that 0Bur

is perpendicular to the plane of the loop.

The magnetic force acting on the loop is : (1983; 1M)

(a) ir 0Bur

(b) 2pir 0Bur

(c) zero (d) pir 0Bur

3. A rectangular loop carrying a current i is situated neara long straight wire such that the wire is parallel to oneof the sides of the loop and is in the plane of the loop.If steady current I is established in the wire as shownin the figure, the loop will : (1985; 2M)

i

i

(a) rotate about an axis parallel to the wire(b) move away from the wire(c) move towards the wire(d) remain stationary

4. Two thin long parallel wires separated by a distance bare carrying a current i ampere each. the magnitude ofthe force per unit length exerted by one wire on theother is (1986; 2M)

(a)2

02

i

b(b)

20

2ibp

(c)0

2ibp (d)

02

2

i

bp

5. Two particles X and Y having equal charges afterbeing accelerated through the same potential difference,enter a region of uniform magnetic field and describecircular paths of radii R1 and R2 respectively. The ratioof the mass of X to that of Y is : (1988; 2M)(a) (R1/R2)

1/2 (b) R2/R1(c) (R1/R2)

2 (d) R1/R26. A current I flows along the length of an infinitely

long, straight, thin-walled pipe. Then : (1993; 2M)(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.

7. A uniform magnetic field with a slit system as shown in figure is to be used as a momentum filter for high energycharged particles with a field B Tesla, it is found that the filter transmits a-particles each of energy 5.3 MeV. Themagnetic field is increased to 2.3 B Tesla and deutrons are passed into the filter. The energy of each deuterontransmitted by the filter is .... MeV. (1997C; 1M)

Source Detector

B

TRUE/FALSE1. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field.

(1981; 2M)2. There is no change in the energy of a charged particle moving in magnetic field although a magnetic force is acting

on it. (1983; 2M)3. A charged particle enters a region of uniform magnetic field at an angle of 85 to the magnetic line of force. The

path of the particle is a circle. (1983; 2M)4. An electron and proton are moving with the same kinetic energy along the same direction. When they pass through

a uniform magnetic field perpendicular to the direction of their motion, they describe circular path of the same radius.(1985; 3M)

278

7. A battery is connected between two points A and Bon the circumference of a uniform conducting ring ofradius r and resistance R. One of the arcs AB of thering subtends an angle q at the centre. the value of themagnetic induction at the centre due to the current inthe ring is : (1995; 2M)(a) proportional to (180 q)(b) inversely proportional to r(c) zero, only if (q = 180)(d) zero for all values of q

8. A proton, a deutron and an a-particle having the samekinetic energy are moving in circular trajectories in aconstant magnetic field. If rp , rd and ra denoterespectively the radii of the trajectories of theseparticles, then : (1997; 1M)(a) ra = rp < rd (b) ra > rd > rp(c) ra = rd > rp (d) rp = rd = ra

9. Two particles, each of mass m and charge q, areattached to the two ends of a light rigid rod of length2R. The rod is rotated at constant angular speed abouta perpendicular axis passing through its centre. Theratio of the magnitude of the magnetic moment of thesystem and its angular momentum about the centre ofthe rod is : (1995; 2M)(a) q/2m (b) q/m(c) 2q/m (d) q/pm

10. Two very long straight parallel wires carry steadycurrents I and I respectively. The distance betweenthe wires is d. At a certain instant of time, a pointcharge q is at a point equidistant from the two wires

in the plane of the wires. Its instantaneous velocity vr

is perpendicular to this plane. This magnitude of theforce due to the magnetic field acting on the charge atthis instant is : (1998; 2M)

(a)0 Iqv2 dp (b)

0 Iqvdp

(c)02 Iqvdp (d) zero

11. A charged particle is released from rest in a region ofsteady and uniform electric and magnetic fields whichare parallel to each other. The particle will move in a:

(1999; 2M)(a) straight line (b) circle(c) helix (d) cycloid

12. a circular loop of radius R, carrying current I, lies in x-y plane with its centre at origin. The total magnetic fluxthrough x-y plane is : (1999; 2M)(a) directly proportional to I(b) directly proportional to R

(c) inversely proportional to R(d) zero

13. An infinitely long conductor PQR is bent to form aright angle as shown in figure A current I flowsthrough PQR, the magnetic field due to this current atthe point M is H1. Now, another infinitely long sraightconductor QS is connected at Q, so that current is I/2 in QR as well as in QS, the current in PQ remainingunchanged. The magnetic field at M is now H2. Theratio H1/H2 is given by : (2000; 2M)

90

90Q

M

P

R

S

(a) 1/2 (b) 1(c) 2/3 (d) 2

14. An ionized gas contains both positive and negativeions. If it is subjected simultaneously to an electricfield along the +x-direction and a magnetic field alongthe +z-direction, then : (2000; 2M)(a) positive ions deflect towards + y-direction and

negative ions towards y direction(b) all ions deflect towards + y-direction(c) all ions deflect towards y-direction(d) positive ions deflect towads y direction and

negative ions towards + y directions

15. A particle of charge q and mass m moves in a circularorbit of radius r with angular speedh w. The ratio ofthe magnitude of its magnetic moment to that of itsangular momentum depends on : (1997; 2M)(a) w and q (b) w, q and m(c) q and m (d) w and m

16. Two long parallel wires are at a distance 2d apart. Theycarry steady equal currents flowing out of the plane ofthe paper as shown. The variation of the magnetic fieldB along the line XX' is given by : (2000; 2M)

d d

x xx' x''

d d

(a) (b)

279

d

d

x' x'x xd

d

(c) (d)

17. A non-planar loop of conducting wire carrying a currentI is placed as shown in the figure. Each of the straightsections of the loop is of length 2a. The magnetic fielddue to this loop at the point P (a, 0, a) points in thedirection : (2001; 2M)

x

y

z

(a) $1 ( )2

+j k$ (b)$1 ( )

3+ +j k i$ $

(c)$1 ( )

3+ +i j k$ $ (d) $

1( )

2+i k$

18. A coil having N turns is wound tightly in the form ofa spiral with inner and outer radii a and b respectively.When a current I passes through the coil, the magneticfield at the centre is : (2001; 2M)

(a) 0 NIb

(b) 02 NI

a

(c)0

2( )NI

a b in ba (d)

02( )

NIb a In

ba

19. Two particles A and B of masses mA and mB respectivelyand having the same charge are moving in a plane. Auniform magnetic field exists perpendicular to thisplane. The speeds of the particles are vA and vBrespectively and the trajectories are as shown in thefigure. Then : (2001; 2M)

B

A

(a) mAvA < mBvB(b) mAvA > mBvB(c) mA < mB and vA < vB(d) ma = mb and vA = vB

20. a long straight wire along the z-axis carries a currenti in the negative z-direction. the magnetic vector field

Bur

at a point having coordinate (x, y) on the z = 0 planeis : (2001; 1M)

(a)0

2 2( )

2 ( )

I y x

x yp +

i j$ $(b)

02 2

( )

2 ( )

I x y

x y

+

p +

i j$ $

(c)0

2 2( )

2 ( )

I x y

x yp +

j i$ $(d)

02 2

( )

2 ( )

I x y

x yp +

i j$ $

21. The magnetic field lines due to a bar magnet arecorrect shown in : (2002; 2M)

(a)

N

(b)

N

S

(c)

N

S

(d)

N

S

22. A particle of mass m and charge q moves with aconstant velocity v along the positive x-direction. Itenters a region containing a uniform magnetic field Bdirected along the negative z-direction, extending fromx = a to x = b. The minimum value of v required so thatthe particle can just enter the region x > b is :

(2002; 2M)

(a)qbBm (b)

( )q b a Bm

(c)qaBm (d)

( )2

q b a Bm

+

23. For a positively charged particle moving in a x-y planeinitially along the x-axis, there is a sudden change inits path due to the presence of electric and/or magneticfields beyond P. The curved path is shown in the x-y plane and is found to be non-circular.

P xO

y

Which one of the following combinations is possible?(2003; 2M)

280

(a) 0;=Eur $b c= +B j k

ur $ (b) ;a=E iur $ $c a= +B k i

ur$

(c) 0;=Eur $c b= +B j k

ur $ (d) ;a=E iur $ $c b= +B k j

ur $

24. A conducting loop carrying a current I is placed in auniform magnetic field point into the plane of the paperas shown. The loop will have a tendency to :

(2003; 2M)

YB

X

I

+

(a) contract(b) expand(c) move towards + ve x-axis(d) move towards -ve x-axis

25. A current carrying loop is placed in a uniform magneticfield in four different orientations, I, II, III and IV,arrange them in the decreasing order of potentialenergy : (2003; 2M)

n^ B

(I)

B

n^(II)

n^

B

(III)

n^

B

(IV)

(a) I > III > II > IV (b) I > II > III > IV(c) I > IV > II > III (d) III > IV > I > II

26. An electrons moving with a speed u along the positivex-axis at y = 0 enters a region of uniform magnetic field

$0B=B k

urwhich exists to the right of y-axis. The

electrons exits from the region after sometimes with thespeed v at co-ordinate y, then : (2004; 2M)

e u

y

x

(a) v > u, y < 0 (b) v = u, y > 0(c) v > u, y > 0 (d) v = u, y < 0

27. A magnetic field $0B=B Jur

exists in the region a < x

< 2a and $0B=B Jur

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

B0 is a positive constant. A positive point charge

moving with a velocity 0v=v ir $ where v0 is a positive

constant, enters the magnetic field at x = a. Thetrajectory of the charge in this region can be like :

(2007; 3M)

B0

B0

0 a 2a 3ax

(a) a 2a 3ax

z

(b)x

a 2a 3a

z

(c)x

a 2a 3a

z

(d)a 2a 3a x

z

OBJECTIVE QUESTIONS

More than one options are correct?

1. A proton moving with a constant velocity passesthrough a region of space without any change in itsvelocity. If E and B represent the electric and magneticfields respectively, this region of space may have :

(1985; 2M)(a) E = 0, B = 0 (b) E = 0, B 0(c) E 0, B = 0 (d) E 0, B 0

2. A particle of charge + q and mass m moving under the

influence of a uniform electric field E i$ and uniform

magnetic field B $k follows a trajectory from P to Q as

shown in figure. The velocities at P and Q are v i$ and

2j$ . Which of the following statement (s) is/arecorrect? (1991; 2M)

281

2a 2VQ

x

a

Py

v E

B

(a)23

4mv

Eqa

=

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

334

mva

(c) Rate of work done by the electric field at P is zero(d) Rate of work done by both the fields at Q is zero

3. H+, He+ and O2+ all having the same kinetic energypass through a region in which there is a uniformmagnetic field perpendicular to their velocity. Themasses of H+, He+ and O2+ are 1 amu, 4amu and 16 amurespectively. Then : (1999; 3M)(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

4. Which of the following statement is correct in thegiven figure. (2006; 5M)

B C

D

OO

A

l2l1

infinitely long wire kept perpendicularto the paper carrying current inwards

(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

5. A particle of mass m and charge q, moving withveloicty v enters Region II normal to the boundary asshown in the figure. Region II has a uniform magneticfield B perpendicular to the plane of the paper. Thelength of the Region II is l. Choose the correct choice(s). (2008; 4M)

(a) The particle enters Region III only if its velocity

mqlB

v >

(b) The particle enters Region III only if its velocity

mqlB

v P

mq

\ ra > rpor track B is of a - particle.

2. Let R be the radius of circle. Then,

2pR = L or R = 2Lp

288

3. Equivalent current i = qfand magnetic moment m = (ipr2) = pqfr2

Substituting the values, we haveM = (p) (0.5 1010)2 (1016) (1.6 1019)= 1.26 1023 A-m2

4. At C magneic field due to wires PQ and RS will bezero.Due to wire QR,

B1 0 0

1 1

12 2 4

I IR R

= =

(perpendicular to paper outwards)

And due to wire SP,

B2 0 02 2

12 2 4

I IR R

= =

(perpendicular to paper inwards)

\ Net magnetic field would be,

B 01 2

1 14

IR R

=

(perpendicular to paper outwards)

5. We can complete the loop EDCBE by assuming equaland opposite current I in wire BE.

x

z

yC

D

E

B AF

BNet force on loop EDCBE will be zero. Similarly, forceon wires FE and BA is also zero, because these areparallel to magnetic field.So, net force is only on wire EB.

Fur

= I [(L i$ ) (B j$ )]

= ILB $k\ Magnitude of force is ILB and direction of force ispositive z.

6.

x

z

yE

A

C D

BF H

I

G

In metal charge carriers are free electrons. Current is inpositive direction of x-axis. Therefore, charge carrierswill be moving in negative direction of x-axis.

mFur

= q ( Vur

Bur

)

= (e) (V i$ ) (B j$ )

nFur

= eVB $kDue to this magnetic force, electrons will be collectingat face ABCD, therefore, lowering its potential.

7. Radius of circular path is given by

R =2mv P Km

Bq Bq Bq= =

Radius in both the cases are equal. Therefore,

2K m

Bqa a

a=

22.3

d d

d

K mBq ,

dqqa

12 2ee

= = and d

mm

a 42

2= =

\ Kd =2

2.3 .dd

q m Kq m

aa

a

Kd =21

2.3 (2)(5.3)2

MeV

Kd = 14.0185 MeV

TRUE/FALSE

1. A current carrying coil is a magnetic dipole. Netmagnetic force on a magnetic dipole in uniform field iszero.

2. Magnetic force acting on a charged particle is alwaysperpendicular to its velocity or work done by a magneticforce on a charged particle is always zero. Hence, amagnetic force cannot change the energy of chargedparticle.

3. The path will be a helix. Path is circle when it entersnormal to the magnetic field.

4.2km

rBq

=

or r m (k , q and B are same)mp > me \ rp > re

OBJECTIVE QUESTIONS (ONLY ONE OPTION)

1. In non-uniform magnetic field the needle will experienceboth a force and a torque.

2. Magnetic force on a current carrying loop in uniformmagnetic field is zero.

3. Straight wire will produce a non-uniform field to the

right of it. bcFur

and dcFur

will be calculated by integrationbut these two forces will cancel each other. Further

M = iA = ipR2 2

4L i=

p

289

force on wire ab will be towards the long wire and onwire cd will be away from the long wire. But since thewire ab is nearer to the long wire, force of attractiontowards the long wire will be more. Hence, the loop willmove towards the wire.

b c

da

I

\ (c)

4. Force per unit lenght between two wires carryingcurrents i1 and i2 at distance r is given by :

Fl =

02p

1 2i ir

Here, i1 = i2 = i and r = b

\Fl =

20

2ibp

\ (b)

5. R =2qVmBq

or R m

or1

2

RR =

X

Y

mm

orX

Y

mm =

21

2

RR

\ (c)

6. Using Ampere's circuital law over a circular loop ofany radius less than the radius of the pipe, we can seethat net current inside the loop is zero. Hence, magneticfield at every point inside the loop will be zero.

7. For a current flowing into a circular arc, the magneticinduction at the centre is

q

l2

l1

A O

B

B = 04

ir

p

q

or B iqIn the given problem, the total current is divided intotwo arcs

i 1

resistanceof arc

1

lengthofarc

1

anglesubstendedatcentre( )q

or iq = constanti.e., magnetic field at centre due to arc AD is equal andopposite to the magnetic field at centre due to arcACD. Or the net magnetic field at centre is zero.

8. Radius of the circular path is given by

r =mVBq Bq

km2=

Here, K is the kinetic energy of the particle.

Therefore, r mq

if K and B are same.

\ rp : rd : ra 1 2 4

: : 1: 2 :11 1 2

= =

Hence ra = rp < rd9. Current i = (frequency) (charge)

= (2 )2q

w p

=qwp

w

(q,m) (q,m)RR

Magnetic moment, M = (i) (A)

=2( )

qR

w p p = (qwR2)

Angular momentum, L = Iw = 2 (mR2)w

\ML

=2

22( )

q R

mR

w

w 2qm

=

10. Net magnetic field due to both the wires will bedownward as shown in the figure.

290

x

zy

dB

v

II

Since, angle between vr

and Bur

is 180.Therefore, magnetic force

mFur ( ) 0q v= =B

r ur

11. The charged particle will be accelerated parallel (if it isa positive charge) or antiparallel (if it is a negativecharge) to the electric field, i.e., the charged particlewill move parallel or antiparallel to electric and magneticfield. Therefore, net magnetic force on it will be zeroand its path will be a straight line.

12. Total magnetic flux passing through whole of the X-Yplane will be zero, because magnetic lines from aclosed loop. So, as many lines will move in -Z directionsame will return to + Z direction from the X-Y plane.

13. H1 = Magnetic flux at M due to PQ + Magnetic fieldat M due to QRBut magnetic field at M due to QR = O\ Magnetic field at M due to PQ (or due to currentI in PQ) = H1Now H2 = Magnetic field at M due to PQ (current I)+ magnetic field at M due to QS (current I/2) +(magnetic field at M due to QR

= 11 02H

H + +

= 132

H

1

2

HH =

23

Note : Magnetic field at any point lying on the currentcarrying straight conductor is zero.

B = 0

i

14. We can write iEE .=r

and $B=B kur

Velocity of the particle will be along the direction of

Er

.Therefore, we can write

Vur

= AqEi$

In Eur

, Bur

and Vur

, A, E and B are positive constantwhile q can be positive or negative.Now, magnetic force on the particle will be

mFuuur

= q ( )V Bur ur

= q {AqE i$ } {B $k }

= q2 AEB ( i$ $k )

mFuuur

= q2 AEB ( j$ )

Since, mFuuur

is along negative y-axis, no matter what is

the sign of charge q. Therefore, all ions will deflecttowards negative y-direction.

15. Ratio of magnetic moment and angular momentum isgiven by

ML

= 2qm

which is a function of q and m only. This can bederived as follows :

M = i A= (q f). (pr2)

= (q) ( )22 rw p p

=2

2q rw

and L = Iw= (mr2w)

\ML

=

2

22r

q

mr

w

w

= 2qm

16. If the current flows out of the paper, the magnetic fieldat points to the right of the wire will be upwards andto the left will be downwards as shown in figure.

B

B

B

291

Now, let us come to the problem,Magnetic field at C = 0Magnetic field in region BX' will be upwards (+ve)because all points lying in this region are to the rightof both the wires.

XXA C B

Magnetic field in region AC will be upwards (+ve),because points are closer to A, compared to B. Similarlymagnetic field in region BC will be downwards (ve).Graph (b) satisfies all these conditions. Therefore,correct answer is (b).

17. The magnetic field at P (a, 0, a) due to the loop isequal to the vector sum of the magnetic field producedby loops ABCDA and AFEBA as shown in the figure.

kj

i

^^

^

P (a, 0, a)

C

B

FA

D

E

Magnetic field due to loop ABCDA will be along i$

and due to loop AFEBA, along $k . Magnitude ofmagnetic field due to both the loops will be equal.Therefore, direction of resultant magnetic field at P will

be $( )12 +i k$

18. Consider an element of thickness dr at a distance rfrom the centre. The number of turns in this element.

dN = N

drb a

Magnetic field due to this element at the centre of thecoil will be

drb

ra

dB = 0 ( )

2dN Ir

= 0 .2

I N drb a r

\ B =r b

r adB

=

=0

2( )NI

b a= ln

ba

19. Radius of the circle mvBq

=

or radius mv if B and q are same.(Radius)A > (Radius)B

\ mAvA > mBvB

20. Magnetic field at P is Bur

, perpendicular to OP in thedirection shown in figure.

O

P (x, y)r

y

qq

i x

B

So, B = B sin q i$ Bcosq j$

Here, B =2

Irp

0

sin qyr

= and cos q xr

=

\ Bur 0

21

.2 I

r=

p)( jxiy -

)(2)(

220

yxjxiyI

+p-m=

(as r2 = x2 + y2)

21. Magnetic lines form closed loop. Inside magnet theseare directed from south to north pole.

22. If (b a) r(r = radius of circular path of particle)The particle cannot enter the region x > bSo, to enter in the region x > b

r > (b a)

or ( )mv

b aBq

> or( )q b a Bv

m>

23. Electric field can deviate the path of the particle in theshown direction only when it is along negative y-

direction. In the given options Eur

is either zero oralong x-direction. Hence, it is the magnetic field whichis really responsible for its curved path. Options (a)and (c) cannot be accepted as the path will be helix inthat case (when the velocity vector makes an angle

292

other than 0, 180 or 90 with the magnetic field, pathis a helix) option (d) is wrong because in that casecomponent of net force on the particle also comes in

$k direction which is not acceptable as the particle ismoving in x-y plane. Only in option (b) the particle canmove in x-y plane.

24. Net force on a current carrying loop in uniformmagnetic field is zero. Hence, the loop cannot translate.So, options (c) and (d) are wrong. From Fleming's lefthand rule we can see that if magnetic field isperpendicular to paper inwards and current in the loop

is clockwise (as shown) the magnetic force mFur

oneach element of the loop is radially outwards. or theloops will have a tendency to expand.

Fm

y

x

25. U = BMrr

. = MB cos qHere, = magnetic moment of the loop

q = angle between Mr

and Br

U is maximum when q = 180 and minimum when q = 0So, as q decrease from 180 to 0 its PE also decrease.

26. Magnetic force does not change the speed of chargedparticle. Hence v = u. Further magnetic field on theelectron in the given condition is along negative y-axisin the starting. Or it describes a circular path inclockwise direction. Hence, when it exists from thefield, y < 0.Therefore, the correct option is (d).

27. ( )m q= F v Bur r ur

Correct option is (a).

OBJECTIVE QUESTIONS (MORE THAN ONE OPTION)

1. If both E and B are zero, then eFur

and mFur

both are

zero. Hence, velocity may remain constant. Therefore,option (a) is correct.If E = 0, B 0 but velocity is parallel or antiparallel to

magnetic field then also eFur

and mFur

both are zero.

Hence, option (b) is also correct.

If E 0, B 0 but eFur

+ mFur

= 0, then also velocity

may remain constant or option (d) is also correct.

2. Magnetic force does not do work. From work-energytheorem :

eFW = DKE

or (qE) (2a) =12

m [4v2 v2]

or E =23

4mvqa

\ Correct option is (a).At P, rate of work done by electric field

= eFur

. vr

= (qE) (v) cos 0

=2 33 3

4 4mv mv

q vqa a

=

Therefore, option (b) is also correct.Rate of work done at Q :

of electric field = eFur

. vr

= (qE) (2v) cos 90 = 0and of magnetic field is always zero. Therefore, option(d) is also correct.

Note that eFur

= qE i$

3. r =2mv p Km

Bq Bq Bq= =

i.e., r mq

if K and B are same.

i.e., +Hr : +Her : 2+Or 1 4 16

: :1 1 2

=

= 1 : 2 : 2Therefore, He+ and O2+ will be deflected equally butH+ having the least radius will be deflected most.

4. BAFur

= 0, because magnetic lines are parallel to thiswire.

CDFur

= 0 because magnetic lines are antiparallel to this

wire.

CBFur

is perpendicular to paper inwards and ADFur

is

perpendicular to paper outwards. These two forces

293

(although calculated by integration) cancel each otherbut produce a torque which tend to rotate the loop inclockwise direction about an axis OO'.

5. Bvrr

^ in region II. Therefore, path of particle is circlein region II. Particle enters in region III if, radius of

circular path, r > l or lBqmv > or m

Bqlv >

If ,mBqlv = ,lBq

mvr == particle will turn back and path

length will be maximum. If particle returns to region I,

,2 Bq

mTt

p== which is independent of v.

v = Bqlm

\ Correct option is (a), (c) and (d).

SUBJECTIVE QUESTIONS

1. Electron pass undeviated. Therefore,

eFuur

= mFuuur

or eE = eBv

or /E V dB

v v= = (V = potential difference between

the plates)

or B =Vdv

Substituing the values, we have

B = 3 6600

3 10 2 10 = 0.1T

Further, direction of eFuur

should be opposite of mFuuur

.

or e Eur

e ( vr

Bur

)

\ Eur

vr

Bur

Here, Eur

is in positive x-direction.

There vr

Bur

should be in negative x-direction or Bur

should be in negative z-direction (or perpendicular) topaper inwards, because velocity of electron is inpositive y-direction.

2. eFuur

= q Eur

= (1.6 1019) (102.4 103) $k

= (1.6384 1016) $k N

mFuuur

= q ( vr

Bur

) = 1.6 1019 (1.28 106 i$ 8 10

2 j$ )

= 1.6384 1016 $k N

Since eFuur

+ mFuuur

= 0

\ Net force on the charged particle is zero, particle willmove undeviated.In time t = 5 106s, the x-coordinate of particle willbecome,x = vxt = (1.28 10

6) (5 106) = 6.4 mwhile y and z-coordinates will be zero.At t = 5 106s electric field is switched off. Onlymagnetic field is left which is perpendicular to itsvelocity. Hence, path of the particle will now becomecircular. Plane of circle will be perpendicular to magneticfield i.e., x-z. Radius and angular velocity of circularpath will become :

mvr

Bq=

26 6

2 19(10 )(1.28 10 )

(8 10 )(1.6 10 )

=

= 1m

Bqm

w = 2 19

26(8 10 )(1.6 10 )

(10 )

= = 1.28 106 rad/s

In the remaining time i.e., (7.45 5) 106 = 2.45 106sAngle rotated by particle,q = wt = (1.28 106) (2.45 106) = 3.14 rad = 180\ z -coordinates of particle will become :

z = 2r = 2m

r

P v x

v

zQ

P = (6.4m, 0, 0)Q = (6.4m,0, 2m)

while y-coordinate will be zero.\ Position of particle at t = 5 105s is P = (6.4 m, 0,0)and at t = 7.45 106 s is Q = (6.4 m, 0.2m)

3. Inside a magnetic field speed of charged particle doesnot change. Further, velocity is perpendicular tomagnetic field in both the cases hence path of theparticle in the magnetic field will be circular. Centre ofcircle can be obtained by drawing perpendicular tovelocity (or tangent to the circular path at E and F.Radius and angular speed of circular path would be

r =mvBq

294

and w =Bqm

45

45

r

r

45G

F

v

E

v

q

C

45

45 F

C

E

90

(i) Refer figure (a) :CFG = 90 q and CEG = 90 45 = 45Since, CF = CE\ CFG = CEGor 90 q = 45or q = 45Further, FG = GE = r cos 45

\ EF = 2FG = 2r cos 45 2 cos45mv

Bq

=

=

27 7

19

12(1.6 10 )(10 )

2(1)(1.6 10 )

= 0.14m

Note : That in this case particle completes 14

th of circle

in the magnetic field.

(ii) Refer figure (b) : In this case particle will complete

34

th of circle in the magnetic field. Hence, the time

spent in the magnetic field :

t =34

(time period of circular motion)

3 2 34 2

m mBq Bq

p p= =

27

19(3 )(1.6 10 )

(2)(1)(1.6 10 )

p =

= 4.712 108s

4. (i) sinmv

rBq

q=

27 5

191.67 10 )(4 10 )(sin60 )

(0.3)(1.6 10 )

( =

= 1.2 102m

(ii) 2

( cos )m

p vBq

p= q

27 5

19(2 ) 1.67 10 )(4 10 )(cos60 )

(0.3)(1.6 10 )

p ( =

= 4.37 102m

5. (a) Direction of current at B should be perpendicularto paper outwards. Let current in this wire be iB. Then,

0102 211

A ip +

= 0

2 (10/11)B i

p

or B

A

ii

= 1032

90

B2

B1

S

B

A

or iB = 1032 A

i 10 9.632

= = 3A

(b) Since, AS2 + BS2 = AB2

\ ASB = 90At S : B1 = Magnetic field due to iA

02 1.6

Ai=p

7(2 10 )(9.6)1.6

= = 12 107T

B2 = Magnetic field due to iB

02 1.2

Bi=p

7(2 10 )(3)1.2

= = 5 107T

Since, B1 and B2 are mutually perpendicular. Netmagnetic field at S would be :

B 2 21 2B B= + 7 2 7 2(12 10 ) (5 10 )= + = 13 107 T

(c) Force per unit length on wire B :

Fl =

02

A Bi irp (r = AB = 2m)

=7(2 10 )(9.6 3)

2

= 2.88 106 N/m

6. Magnetic field at O due to L and M is zero. Due to Pmagnetic field at O is :

750

11 (10 )(10)

T 5.0 10 T2 2 0.02

iB

OR = = = p

295

(perpendicular to paper outwards)Similarly, field at O due to Q would be,

750

21 (10 )(10)

T 5.0 10 T2 2 0.02

iB

OS = = = p

(perpendicular to paper outwards)Since, both the fields are in same direction, net fieldwill be sum of these two.\ Bnet = B1 + B2 = 10

4TDirection of field is perpendicular to the paper outwards.

7. (a) Magnetic field at R due to both the wires P and Qwill be downwards as shown in figure. Therefore, netfield at R will be sum of these two.

P Q R

BP

BQ

v

B = BP + BQ

= 0 0 2 5 2 2

QP II +p p

=0 2.5

2 5 2I + p

= ( )0 14

I +p

= 107 (I + 1)Net force on the the electron will be,

Fm = Bqv sin 90

M NR

1m 1mor (3.2 1020) = (107) (I + 1) (1.6 1019) (4 105)or I + 1 = 5\ I = 4A

(b) Net field at R due to wires P and Q isB = 107 (I + 1)T

= 5 107 T (I = 4A)Magnetic field due to third wire carrying a current of2.5 A should be 5 107T in upwards direction so, thatnet field at R becomes zero. Let distance of this wirefrom R be r. Then,

0 2.52 rp = 5 10

7

or7(2 10 )(2.5)r

= 5 107m

or r = 1 m

So, the third wire can be put at M or N as shown infigure.If it is placed at M, then current in it should beoutwards and if placed at N, then current be inwards.

8. (a) Magnetic field at P due to arc of circle.Subtending an angle of 120 at centre would be :

M

I

N

y

y v

x

x60

60

60r

a

+QP

B1 =13 (field due to circle)

=01

3 2I

a

aI

60m= (outwards)

00.16 Ia

= (outwards)

or 1Bur

= $00.16 I

ak

Magnetic field due to straight wire NM at P :

B2 =0

4Irp (sin 60 + sin60)

Here, r = a cos 60

\ B2 =0

2p cos60

Ia (2 sin 60)

or B2 =0

4pIa tan 60

=00.27 I

a (inwards)

or 2Bur

= $00.27 I

ak

\ netBur

= 1Bur

+ 2Bur $00.11 I

a= k

Now, velocity of particle can be written as,

vr

= v cos 60 i$ + v sin 60 j$ 32

v v2

+= i j$ $

Magnetic force

296

mFuuur

Q= (vB)r ur 00.11

2IQv

a= j$ 0

0.11 32

IQva

i$

\ Instantaneous acceleration

ar

mmF=

uuur

( )00.11 32 IQvam

= j i$ $

(b) In uniform magnetic field, force on a current loop iszero. Further, magnetic dipole moment of the loop will

be, Muur

= (IA) $kHere, A is the area of the loop.

( )21 13 2A a= p [2 a sin 60] [a sin60]2 2

3 2a ap= sin 120

= 0.61a2

\ Muur

= (0.61 Ia2) $k

Given, Bur

= B i$

\rt = M

uur B

ur = (0.61 Ia2 B) j$

9. Let us assume a segment of wire OC at a point P, adistance x from the centre of length dx as shown infigure. Magnetic field at P due to current in wires Aand B will be in the directions perpendicular to AP andBP respectively as shown.

|Bur

| = pm2

0 IAP

Therefore, net magnetic force at P will be along negativey-axis shown

B

y

O

A

90

90IC

x

BB

BA

Bnet = 2 |Bur

| cos q

=02

2 p

IAP

xAP

Bnet =0

p 2.

( )

I x

AP

Bnet =0p 2 2( )

Ix

a x+Therefore, force on this element will be

(F = ilB)

dF = I 0

2 2 Ix

dxa x

p +

(in negative z-direction)

\ Total force on the wire will be

20

2 20 0

x L Lx

I xdxF dF

x a

=

== =

p +

202

I=p

In 2 2

2L a

a

+

(in negative z-axis)

Hence, Fur

p

m-=

2

20I In

2 2

2L a

a

+

$k

10. Kinetic energy of electron, K = 12

mv2 = 2 keV

\ Speed of electron, v 2Km

=

B

v60

S

G

v =16

312 2 1.6 10

9.1 10

= 2.65 107 m/s

Since, the velocity ( )vr

of the electron makes an angle

of q = 60 with the magnetic field Bur

, the path will bea helix. So the particle will hit S if

GS = npHere n = 1, 2, 3 ......

p = pitch of helix 2 mqBp

= v cos q

But for B to be minimum, n = 1

Hence, GS = p = 2 mqBp

v cos q

B = Bmin 2 cos

( )mvq GS

p q=

Substituting the values, we have

Bmin

31 7

19

1(2 )(9.1 10 (2.65 10 )2

(1.6 10 )(0.1)

p =

T

or Bmin = 4.73 103T

297

11. Let m be the mass per unit length of wire AB. At aheight x above the wire CD, magnetic force per unitlength on wire AB will be given by

Fm

Fg

i1=20AA

C

B

D

x=d=0.01m

Fm =0 1 2

2i ixp (upwards) ..(1)

Weight per unit length of wire AB isFg = mg (downwards)

Here, m = mass per unit length of wire ABAt x = d, wire is in equilibrium i.e.,

Fm = Fg or 0 1 2

2i idp = mg

or0 1 2

22

i i

dp=

mgd ...(2)

When AB is depressed, x decreased therefore, Fm willincrease, while Fg remains the same. Let AB isdisplaced by dx downwards. Differentiating Eq. (1)w.r.t. x we get

dFm =0 1 2

22

i i

xp.dx ...(3)

i.e. restoring force, F = dFm dxHence, the motion of wire is simple harmonic.From Eqs. (2) and (3), we can write

dFm = mgd

.dx (x = d)

\ Acceleration of wire a .g

dxd

= Hence, period of oscillation

T 2dxa

= p|displacement|

=2p|acceleration|

or 2d

Tg

= p 0.0129.8

= p or T = 0.2 s

12. (i) In ground state (n = 1) according to Bohr's theory:

mvR 2h=p

or v =2

hmRp

Now, time period, T 2 R

vp= 2

/ 2R

h mRp=p

2 24 mRh

p=

Magnetic moment M = iA

where i charge

=timeperiod 2 2

e=

4 mRh

p 2 24

eh

mR=

p

and A = pR2

\ M = (pR2) 2 24

eh

mR

p

or M = 4eh

mp

Direction of magnetic moment Muur

is perpendicular tothe plane of orbit.

(ii) tr

= Muur

Bur

Muur

= MB sin q

where q is the angle between Muur

and Bur

q = 30

\ t = 4eh

m p

(B) sin 30

\ t = 8ehB

mp

The direction of tr

is perpendicular to both Muur

and

Bur

.

13. (a) Magnetic field will be zero on the y-axis i.e.,x = 0 = z

z-axisi

x = d O x = + dx

IVIIIIII

y

Magnetic field cannot be zero in region I and regionIV because in region I magnetic field will be alongpositive z-direction due to all the three wires, while inregion IV magnetic field will be along negative z-axis

298

due to all the three wires. It can zero only in region IIand III.

x dx

d+x

1 2 3

i ix=x

Let magnetic field is zero on line (z = 0) and x = x. Thenmagnetic field on this line due to wires 1 and 2 will bealong negative z-axis and due to wire 3 along positivez-axis. Thus

B1 + B2 = B3

or0 0

2 2i i

d x x+

p + p =0

2 i

d xp

or1 1

d x x+

+ =1d x

This equation gives x = + 3

d

x = d x = 0 x = dx

32

zy-axis

1

where magnetic field is zero.

(b) In this part we change our coordinate axes system, justfor better understanding.There are three wires 1, 2 and 3 as shown in figure. Ifwe displace the wire 2 towards and z-axis.

F F2

qqr r

3z

d d

1

then force of attraction per unit length between wires(1 and 2) and (2 and 3) will be given by

F =2

02

irp

The components of F along x-axis will be cancelledout. Net resultant force will be towards negative z-axis(or mean position) and will be given by

Fnet =2

02

irp

)cos2( q

=2

022

i zr r

p

Fnet =2

02 2

( )

i zz dp +

If z < < d, then

z2 + d2 = d2 and Fnet = 2

02

.

2i

zd

p

Negative sign imples that Fnet is restoring in nature.Therefore, Fnet zi..e, the wire will oscillate simple harmonically.Let a be the acceleration of wire in this position andl is the mass per unit length of wire then

Fnet = la = 2

02

iz

d

p

or a = 2

02

.

iz

d

pl

\ Frequency of oscillation

f =1

2paccleration

displacement

1=2

azp

=1

2idp

0pl

or 0

2ifd

=p pl

14. j$ =EEur

or 0

0:

B v=

B vi

ur r$

$k =00

0v Bv B

r ur

E and B

v0x

z

299

Force due to electric field will be along y-axis. Magneticforce will not affect the motion of charged particle inthe direction of electric field (or y-axis). So,

ay eF qE

m m= = = constant. Therefore, vy = ayt .

qE tm

=

The charged particle under the action of magnetic field

describes a circle in x-z plane (perpendicular to Bur

)with

T =2 mBqp

or 2 qBT mpw = =

Initially (t = 0) velocity was along x-axis. Therefore,

magnetic force m( )Fur

will be along positive z-axis

( )m 0q = F v Bur r ur

. Let it makes an angle q with x-axis

at time t, then

zvz v0

v0

vx

x

\ vx = v0 cos wt = v0 cos qB

tm

...(2)

vz = v0 sin wt = v0 sin qB

tm

...(3)

From Eq. (1), (2) and (3),

vr

= vx i$ + vy j$ + vz $k

\ vr

= ( )00 cos qB qEv t tm m E +

Evurr

+

BvBvt

mqBv

0

00 sin

rr

or vr

= ( ) ( )0cos qB qt tm m +

v Er ur

+

BBvt

mqB

rr0sin

Note : The path of the particle will be a helix of increasingpitch. The axis of the helix will be along y-axis.

15. Magnetic moment of the loop, $( )iA=M kuur

= $20( )I L k

Magnetic field field, Bur

= (B cos 45) i$ + (B sin 45) j$

= ( )2

B+i j$ $

(a) Torque acting on the loop, t = M Br uur ur

= $20( )I L k ( )2

B +

i j$ $

\ tr

=2

0 ( )2

I L Bj i$ $

or | tr

| = I0L2 B

(b) Axis of rotation coincides with the torque and since

torque is in j$ i$ direction or parallel to QS. Therefore,

the loop will rotate about an axis passing through Qand S as shown in the figure.

S

P Q

R

x

Angular acceleration, a I

t=

r

where I = moment of inertia of loop about QS.IQS + IPR = IZZ

(From theorem of perpendicular axis)But IQS = IPR

\ 2IQS = IZZ 24

3ML=

IQS =22

3ML=

\ a I

t=

r

=2

022 / 3

I L B

ML03

2I BM

=

\ Angular by which the frame rotates in time Dt is

21 ( )2

tq = a D or 203

.( )2

I Bt

Mq = D

300

16. (a) q = 30

sin q =LR

Here R =0

0

mvB q

\ sin 30 =0

0

LmvB q

or12

=0

0

B qLmv

\ L =0

02mvB q

(b) In part (a)

sin 30 =LR

or12

=LR

or L = R/2Now when L' = 2.1 L

or2.12

R

L' > RTherefore, deviation of the particle is q = 180 is asshown.

\ ivv f 0-=r

and tAB = T/20

mB qp

=

17. (a) Magnetic field ( )Bur

at the origin = magnetic field

due to semicircle KLM + Magnetic field due to othersemicircle KNM

\ Bur

= )(4

)(4

00 jRI

iRI m

+-m

Bur

)(4

4

4

000 jiRI

jRI

iRI

+-m

=m

+m

-=

\ Magnetic force acting on the particle

Fur

= ( )q V Bur ur

RI

jiivq4

)}(){( 00m

+--=

kR

IqvF

400m-=

r

(b) = =KLM KNM KMF F Fur ur ur

and KMFur

= B I (2R) i$

= 2BIR i$

1Fur

= 2Fur

= 2 BIR i$

Total force on the loop, 1 2= +F F Fur ur ur

or Fur

= 4 BIR i$

Then ADC AC=F Fur ur

or ADC ( )AC B=F iur $

From this we can conclude that net force on a currentcarrying loop in uniform magnetic field is zero. In thequestion, segments KLM and KNM also form a loopand they are also placed in a uniform magnetic fieldbut in this case net force on the loop will not be zero.It would had been zero if the current in any of thesegments was in opposite direction.

18. (a) Given : i = 10 A, r1 = 0.08 m and r2 = 0.12 m. Straightportions i.e., CD etc., will produce zero magnetic fieldat the centre. Rest eight arcs will produce the magneticfield at the centre in the same direction i.e.,perpendicular to the paper outwards or verticallyupwards and its magnitude is

B = Binner arcs + Bouter arcs

=0

1

12 2

ir

0

2

12 2

ir

+

=0

4 p

(pi) 1 2

1 2

r rr r

+

Substituting the values, we have

B 7(10 )(3.14)(10)(0.08 0.12)

(0.08 0.12)+

= T

B = 6.54 105T (Vertically upward or outward normalto the paper)

(b) Force an ACForce on circular portions of the circuit i.e., AC etc.,due to the wire at the centre will be zero becausemagnetic field due to the central wire at these arcs willbe tangential (q = 180) as shown.Force on CDCurrent in central wire is also i = 10A. Magnetic fieldat P due to central wire.

B = 0 .2

ixp

\ Magnetic force on element dx due to this magneticfield

dF = (i) 0 .

2ixp

. dx 20

2dx

ixp

=

301

(f = ilB sin 90)Therefore, net force on CD is

F =

pm

=p

m=

-

= 23

ln22

2012.0

08.0

202

1

ix

dxidF

rx

rx

Substituting the value, F = (2 107) (10)2 ln (1.5) orF = 8.1 106N (inwards)Force on wire at the centreNet magnetic field at the centre due to the circuit is invertical direction and current in the wire in centre isalso in vertical direction. Therefore, net force on thewire at the centre will be zero. (q = 180). Hence,(i) Force acting on the wire at the centre is zero.(ii)Force on are AC = 0(iii) Foce on segment CD is 8.1 106 N (inwards)

19. Let the direction of current in wire PQ is from P to Qand its magnitude be I

x S

I

Q yP

z

R

The magnetic moment of the given loop is :

Muur

= Iab $kTorque on the loop due to magnetic force is :

1tr

= Muur

Bur

= (Iab $k ) (3 i$ + 4$k ) B0 i$

= 3lab B0 j$

Torque of weight of the loop about axis PQ is :

2tr

= rr

Fur

=$( )2

i k$

amg

=2

jmga

We see that when the current in the wire PQ is from

P to Q, 1tr

and 2tr

are in opposite direction, so theycan cancel each other and the loop may remain inequilibrium. So, the direction of current I in wire PQ isfrom P to Q. Further for equilibrium of the loop :

| 1tr

| = | 2tr

|

or 3IabB0 = 2mga

I =06

mgbB

(b) Magnetic force on wire RS is :

Fur

= I ( lr

Bur

)

= I [(b j$ ) {(3 i$ + 4$k ) B0}]

Fur

= IbB0 (3 $k 4 i$ )

20. In equilibrium,2T0 = mg

or T0 =mg2

Magnetic moment, M = iA wp

= Q2 ( )2pR

t = MB sin 90 2

2w

=BQR

Let T1 and T2 be the tensions in the two strings whenmagnetic field is switched on (T1 > T2).For translational equilibrium of ring is vertical direction,

T1 + T2 = mg ...(2)For rotational equilibrium,

(T1 T2) 2D 2

2w

= t =BQR

or T1 T22

2w

=BQR

..(3)

Solving Eqs. (2) and (3), we have

T1 =2

2 2w

+mg BQR

D

As T1 > T2 and maximum values of T1 can be 03

2T

, we

have

032T

= T0 2

max2

w+

BQRD

02 =

mgT

\ wmax =02

DT

BQR

21. BqqVm

r2

= or m

rq

a

pr

ra

a= p

p

m qm q

1 2 14 1 2

= =

22. (a) t = MB = ki (q = 90)

302

\ k =( )=MB NiA B

i i= NBA

(b) t = k .q = NiAB (k = Torsional constant)

\ p= NiABk 2 (as q = p/2)

(c) t = NiAB

or 0 0

t = t t

dt BNA idt

Iw = BNAQ

or w =BNAQ

I..(1)

At maximum deflection whole kinetic energy (rotational)will be converted into potential energy of spring.

Hence, 212

wI = 2max12

qk

Substituting the value, we get

qmax = 2pBN AQI

23. )53sin37(sinp40 +=

rIBp

+

p

m=

54

53

53

44

0

x

I

R

Q

P

3xr

5x

4x

3753

==

xI

kxI

p48

p487 00

\ k = 7