ANSWER KEY

1. (b)

2. (b)

3. (c)

4. (b)

5. (b)

6. (d)

7. (d)

8. (d)

9. (a)

10. (a)

11. (d)

12. (a)

13. (b)

14. (b)

15. (b)

16. (a)

17. (b)

18. (d)

19. (d)

20. (d)

21. (b)

22. (d)

23. (d)

24. (a)

25. (a)

26. (a)

27. (c)

28. (a)

29. (b)

30. (c)

31. (c)

32. (c)

33. (b)

34. (d)

35. (a)

36. (a)

37. (a)

38. (b)

39. (a)

40. (b)

41. (d)

42. (c)

43. (b)

44. (a)

45. (a)

46. (d)

47. (b)

48. (c)

49. (a)

50. (a)

51. (c)

52. (c)

[EE] CLASSROOM PRACTICE TEST-03EMFT (SOLUTIONS)

53. (b)

54. (b)

55. (d)

56. (d)

57. (d)

58. (d)

59. (a)

60. (b)

61. (a)

62. (d)

63. (c)

64. (a)

65. (a)

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1. (b)

• Gradient is always taken for a scalar quantityand it produces a vector.

• Divergence is always taken for a vectorquantity and it results into a scalar.

• Curl is always taken for a vector quantityand it results into a vector.

So,

.A -Gradient of a scalar- correct

A - Gradient of a vector - Incorrect

. A -Divergence of a vector - correct

V - Curl of a vector - correct

2. (b)

Given, electric potential, 2V 3x y yz

So, Electric field intensity

E

= V

= x y zV V Vˆ ˆ ˆa a ax y x

= 2x y zˆ ˆ ˆ6xya a ay3x z

2y zx1,0, 1

ˆ ˆˆ 3 1 a a6 1 a 010E

= yˆ4a 0

2x y z2, 1,4

ˆ ˆ ˆ6 a a a2 1 130 2 4E

= x y zˆ ˆ ˆ12a 8a a V m

Now, potential V at (2, –1, 4)

2, 1,4V = 23 2 1 1 4

= 12 4 = –8V3. (c)

L s

H d H ds

l (Stoke’s Theorem)

sJ ds I (current through the cross-section)

(flux through the surface S)

4. (b)Magnetic field intensity (H) due to wire 1at a distance d/2 (i.e at point P)

1 2

IP

d

I

1H = Id (downward)

And due to wire 2 at P

2H = Id (upward)

The vector addition of these two fields

H = 1 2IH H

d d

= 0

5. (b)

+Q10cm

20 cm

1 2

As the sphere of radius 10 cm encloses the totalcharge Q. So the concentric sphere of radius 20cm will definitely enclose the total charge Q.According to Gauss’s Law(flux)2 = total charges enclosed = (flux)1 = 20 V.m.

6. (d)

In electrostatics, a perfect conductor has followingcharacteristics :1. No electric field exist within a conductor i.e.

electric field will be only external2. Electric field will be normal to its surface.3. Since E V 0 ,there will be no potential

difference between any points on its surface.

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7. (d)

X

Y

Z

2

3

1

1

0 1

1

Circulation F

around closed path

= F.d

l

= 1 2 3

F.d F.d F.d l l l

For curve (1), Z = 0

So, F = x yˆ ˆxa ya

Then, 1

F.dl = x y x y zˆ ˆ ˆ ˆ ˆxa ya dxa dya dza.

= 0 1

1 0

xdx ydy

= 0 12 2

1 0

x y2 2

= 1 1 02 2

The value of l

F.d will be same for curve (2) andcurve (3). So,

F.dl = 0

8. (d)

Ampere Law is a special case of Biot Savart Law.Using Ampere’s Law, magnetic field intensity forsymmetrical current distributor can be obtainedeasily.

[Note : Gauss Law is a special case of Coulomb’sLaw where charge distribution is symmetrical]

Ampere’s Law states that the line integral of Haround a closed path is the same as the netcurrent enclosed by the path.

i.e. enc.L

H dl I

9. (a)

For an electrostatic field (E) :

S

E.d E ds 0 l

or E 0

Any vector field that satisfies above equationsare called conservative or irrotational field. Inother words, vectors whose line integral does notdepend on the path of integration are calledconservative vector fields. This is true for electric

fields as E.d l represents potential difference in

a closed loop which is definitely zero.Hence, Both statements (i) and (ii) are correctand (ii) is the correct explanation of (i).

10. (a)

At dielectric-dielectric boundary, tangentialcomponents of electric field (Et) are the same ontwo sides of the boundary i.e.

1 2

1 1 2 2

t t

t 1 t t 2 t

E E ... (i)We know, D E

D E and D E

Using 1 2t tE and E in equation (i)

1 2t t

1 2

D D... (ii)

Therefore, tangential component of field density(Dt) undergoes some changeacross the interface.

11. (d)

Applying Gauss law on dielectric-dielectricboundary, we get

1 2n nD D = normal components of D on both

sides of interface

s = charge on interface

If no free charge exist at the interface then, s 0

Therefore, from equation (i), we get

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1 2n n

1 1 2 2

D Dand E E

Normal component of D is continuous across theinterface whereas normal component of E isdiscontinuous.

12. (a)++++++++++++++++++

E

r

According to Gauss’s Law,

E ds =

0

Q

lE 2 r = l

0

E = r

0a

2 r

y

(0,1)

1x

Line-1

Line-2 (0,–1)

0 (0,0)

–

So, electric field at the origin by line-1,

1E = y

0a

2 r

= y0

a2

[ r = 1]

and electric field at the origin by line-2,

2E = y0

a2 r

= y0

a2

[ r = 1]

so, resultant electric field intensity at the originE =

1 2E E

= y0

2 a2

= y0

a

13. (b)

E = dVdx

= 2d (5x 10x 9)dx

= –[10x + 10]= –[10(1) + 10]= – 20 V/m

14. (b)

The electric field is given by—

E = V

= – 80 0.6 a

E = 0.4 ˆ–48 a V/m

D = 0E

D = –48 0.4

0 a C/m2

At = 0.6m,

D

= – 48 0.40.(0.6) a

D = – 0.521 × 10–9 a C/m2

The flux density is constant.

So, the flux = D.S

= – 0.521 × 10–9 × 2 z

= – 0.521 × 10–9 × 2 × 0.6 × 1 = – 1.96 nC

From Gaussis law, emc.Q = = –1.96 nC.

15. (b)

The electric field intensity

E .V

= x0x

0ˆ. e a

a

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E = x0

x0

ˆ. e aa

Energy density, WE = 20

1 E2

WE = 2 2

2 x02

0

1 e2 a

Energy stored in region,

WE = 1 1 d2 2

2 x02

0 0 0 0

1 e dx dy dz2 a

WE = 2 2 2 d

02

0

1 e 1(1) (1)2 ( 2 )a

20 2 d

E 20

W 1 e4a

16. (a)

Since field A

is irrotational

A = 0

A

shold be gradient of some scalar fieldA V , as curl of a gradient is always zero.

17. (b)

The magnetization (M)

is given by

M = mX H

=

m m

0 r 0 m

X H X B1 X

M =

y–7

ˆ3.1 0.4a

4 10 (4.1)

M = 2.41 × 105 ya A/m.

18. (d)

Ampere Law states that

H dl Ienclosed.

H dl = – 30 + 10 =– 20A

–30A is taken because the path taken is oppositeto the direction of magnetic field due to 30Acurrent.

19. (d)

If the potential of cylinder at = a is given as V0,

and that of = b is given as 0 (b > a), then the

potential of cylindrical surface at 1 will begiven by

V = 1o

n(b/ρ )V .

n(b/a)ll

Here, V0 = 100 V

b = 1.2 cm.

a = 0.5 cm.

V = 20 V.

20 = 100 1n(1.2/ρ )n(1.2/0.5)ll

1

1.2n

l = 1

1.20.175 1.19

1 1.01 cm.

20. (d)

V

AQ –Q

1 2

Electric field near plate 2 due to plate 1 will be :

E = 0 0

Q2 2A

Therefore, force on plate Z due to electric fieldwill be :

F

= 2 2 2

0 0

Q C VQE2A 2A

C = 0A

d

F

=

2 20

20

A .V2 A .d

F

= 0 22A V

2d

21. (b)

The current density is given by J = 0v

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as v = a

2

0. ˆJ a A/m .

22. (d)

The net capacitance can be considered as twocapacitances in series.

1

1AC

d 2

and 2

2AC

d 2

C =

21 2 1 2

21 21 2

C C 4 A2AC C dd

C = 1 2

1 2

2Ad

23. (d)

For

16 ˆ0 6 : 2 H 16, H a2

For 4 ˆ6 10 : 2 H 16 12, H a

2

For 10 : 2 H 16 12 4 0, H 0

The best relation between H and P is describedby (d).

24. (a)Since both V1 and V2 satisfy the boundarycondition and Laplace equation, hence bothare correct and the solution is not unique.

25. (a)

dsP

According to Stoke’s theorem, which relates lineintegral to surface integral, H.dl = H .ds

Thus, dsP =

P.dl

26. (a)

The current density is given by

J = H

J

=

x y z

x y z

2

ˆ ˆa a a

2x 2y0zz

= y yx y z2 2

x 2 x 2ˆ ˆ ˆa 0 a (0 0) az zz z

y

2

x 2z z

J

= x z3 22 1ˆ ˆa ax 2yz z

at P (1,5,3)

J

= x z3 22 1ˆ ˆa (1 10) a3 3

x zˆ ˆJ 0.81a 0.11 a

27. (c)The current density due to an infinite sheet isgiven by

k = zI ˆ(a )w – for top strip

k = zI ˆ( a )w

– for bottom strip–y

+I

–I

–z

wx

12 P

na2

x

1l

na

The magnetic field due to an infinite sheet ofcurrent is given by

H = n1 ˆk a2

na = unit normal vector directed from the currentsheet to point of interest.For upper strip,

H = z y x1 I Iˆ ˆ ˆ(a a ) ( a )2 w 2w

For lower strip,

H = z y x1 I Iˆ ˆ ˆa ( a ) ( a )2 w 2w

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H at point P due to both strips will be

H = x xI Iˆ ˆ( a ) 2 a

2w w

B = 00 x

I ˆH aw

Now, differential area of the strip,dS = xˆdydz –a

m = ox x

I ˆ ˆB dS a dydzaw

= x

o

y 0 z 0

I dy dzw

l

m = oI xw

l

L = om xI w

l

Inductance per unit lengthLl = 0

xLw

L x

If xx2

then, LL H/m.2

Hence, option (c) is correct.28. (a)

Image theory is applicable only in static fields.(i) Charge distribution is replaced with the

original charge distribution and its imagecharges.

(ii) The conducting surface is replaced by anequipotential surface.

29. (b)

+

++ +

+

Dn

sConductor

• The tangential component of electric field linesdoes not exist on a conducting surface boundary.

• any conducting surface always supports thenormal component of the electric field suchthat the normal component of electric fluxdensity D

is equal to the magnitude of the

surface charge density S present on theconducting surface.

30. (c)

As we know that for an infinitely long currentcarrying conductor

H = I a2 r

IHr Hence, Rectangular hyperbola.

31. (c)

X

Y

–

+

++++++++++++

Z

According to Gauss’s Law

s

E.ds

= 0

charge enclosed

E.2 R. l = 0

.

l

E = 02 R

This is the electric field intensity by an infinitelylong line charge at a distance R, and its directionwill be radially outward.

So, E at (R, 0, 0) = x0

a2 R

32. (c)E = cos2 a – sin2 a

Edd E

d cos2

sin2 d

Integrating ln = –1ln sin22

+ C

The line passes through P(2, 30°, 0),

ln 2 = –12

ln {sin60°} + C

C = 0.62

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ln = –12

ln sin2 +0.62

2ln + ln sin2 = 0.62 × 2

2 2sin 3.46

33. (b)The electric flux leaving the surface, will be equalto the charge contained within it.

90 0.05

s30 .01

Q . d dz

0.052

–20z

0.016

. d . 5e dz

0.05–20z

0.01

50.08 – .2 6 –20

= –0.0209 × (–0.45)

–39.41 10 nC

9.41pC

34. (d)

Magnetic boundary conditions are

1 2 1 2n n 1 n 2 nB B or H H

Therefore, Bn is continuous ; Hn is discontinuousat boundary.Here, Bn, Hn = normal components of B and H.

Also, H1t = H2t = k

1t 2t

1 2

B B k

Therefore, Bt is discontinuous at boundaryHere, Bt, Ht are tangential componentsK = free current density at interface

35. (a)E = 20 e–5y x ycos5xa – sin5xa

= 20 e–5y cos5xxa – 20 e–5y sin5xxa

y

x

Edy – sin 5xdx E cos 5x

dy – tan5x dx

Integrating dy – tan5xdx

y = 15 ln cos5x C –(i)

As the line represented by eq. (i) passes through

,0.1,23 , so

0.1 = 15 ln

5cos3 + C

C 0.24

The equation of stream line is

1y ln cos5x 0.245

36. (a)

The differential area in za direction is given

by zˆds d d a

I = J. ds

I = 2 0.4

20 0

20 d d0.01

= – 20 2 0.4

20 0

dd0.01

= – 20 0.420

1(2 ) n( 0.01)2 l

= – 20 [ln (0.16 + 0.01) – ln (0.01)]

I = – 20 ln0.17 – 20 n(17)0.01

l

I – 178 A .

37. (a) Since, m = r 1 = 6.5–1 = 5.5

then, magnetisation

M

= mH

M

= x y z5.5 ˆ ˆ ˆ10a 25a 40a

= 2x y zˆ ˆ ˆ55a 137.5a 220a A m

38. (b)

D = E

D =

5510 ˆcos(10 t)a

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DJ

= Dt

= 5

510 ˆcos a10 tt

DJ

= 5

5510 sin 1010 t

DJ

= 51 ˆsin a10 t

The displacement current will flow through thesurface area of the cylinderical capacitor, radiallyoutward or inward.

ID = 5D

1 sin (10 t)2 I.J 2 l

ID = 52 .I sin (10 t)

= – 52 (0.4) sin (10 t)

5DI 0.8 sin (10 t) A .

39. (a)

The two planes are 2x + 3y – 12 = 0and 2x + 3y – 18 = 0Using two boundary conditons, we can write thegeneral potential function as V (x1 y) = A (2x + 3y – 12) +100V (x1 y) = A (2x + 3y – 18) + 0A (2x + 3y –12) + 100 = A (2x + 3y –18)100 = –6A

–100A

6

V(x, y) = –100

6 (2x +3y –18)

= –33.33x –50y + 300 E V

x yE 33.33a 50a

The electric field function is independent ofposition.

40. (b)

Let the required point is 3P 0, y, 0Then,

13R

= x y zˆ ˆ ˆ0 4 a y 2 a 0 7 a

= x y zˆ ˆ ˆ4a y 2 a 7a

and 23R

= x y zˆ ˆ ˆ0 3 a y 4 a 0 2 a

= x y zˆ ˆ ˆ3a y 4 a 2aNow, electric field at P3 X-direction,

Ex =

9

322 20

10 25 44 4 y 2 7

322 2

60 3

3 y 4 2

As Ex = 0, we get

20.48y 13.9y 73.12 0

y = –6.89 or –22.11i.e. P3 = (0, –22.11, 0)

41. (d)

The potential at any point due to thecombination of these three charge distributionsis

V =6 9 9

0 0 0

2 10 8 10 5 10n( ) (z) C4 r 2 2

l

Where, r = distance of the point from (2, 0, 0) = distance of the point from line x = 0, z =4.z = z co-ordinate of the point.at m (0, 0, 5) —

r = 2 22 5 29

= 5 – 4 = 1

z = 5V = 0.

0 =6 9 9

0 00

2 10 8 10 5 10n(1) (z) C2 24 29

l

= –6 9 9

122 10 9 10 5 10 5– 0 – C

2 8.85 1029

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C = – 3.34 × 103 + 1.414 × 103

= – 1.93 × 103

The expression for potential is given by—

V = 6 9 9

3

0 0 0

2 10 8 10 5 10n( ) (z) 1.93 10 .4 r 2 2

l

42. (c)

z

x

y

y = 3m

E+E–

y�a

The electric field intensity at a point due to plane

E = sn

0a

2

3, 3,3E = 8 9

y10 36 10 a

6 2

= yˆ30a V m

43. (b)

Contribution to E

at point (1, 1, –1) due to theinfinite sheet 1, infinite sheet 2, and infinite line3 as shown in figure. Let these are E1, E2 and E3respectively.

x=0,z=2

y

z

3

x

x=2(2,0,0)

y=3

(0,–3,0)

E1 =

9s1

x x x90

10 10ˆ ˆa a 180 a2 102

36

E2 =

9s2

y y y90

15 10ˆ ˆ ˆa a 270 a2 102

36and

E3 =

9L L

x z2 90 0

10 10ˆ ˆ ˆa a a 3a2 2 102 10

36

= x zˆ ˆ18 a 3a

E

= x y zˆ ˆ ˆ162 a 270 a 54 a V m

44. (a)

4 cm

12

34

4 cm

Total potential energy stored,

W = 4

n nn 1

1 q V2

= 1 1 1 2 3 3 4 41 1 1 1q V q V q V q V2 2 2 2

Here, potential at charge ‘1’ is,V1 = 21 31 41V V V

= 0

q 1 1 14 0.04 0.04 0.04 2

= 9 9 1 11.2 10 9 10 20.04 2

= 730.89

Here, V1 = 2 3 4V V V V

So, total potential energy stored.

= 14 qV2

= 92 1.2 10 730.89

= 1.75 J45. (a)

Step 1: Since we wish to determine H at distance

such that

b (b t)

We first draw an amperian loop (L) of radius

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t

b

a yz

x

Step 2: Using Ampere’s Law

enc due toL due to outerinner conductorconductor

H dl H 2 I I J dl (i)

Now,

J is the current density of the outer conductor

and is along za

z22

I ˆJ ab t b

…(ii)

2

enc z z220 b

Iˆ ˆI I a d d ab t b

…(iii)

or

22 2

enc 22 0

1 bI I 1 ( )2 2b t b

or

2 2

enc 2bI I 1

t 2bt

Step 3: using equation (iv) in equation (i)

2 2

2bH 2 I 1

t 2bt

2 2

21 bIH

2 t 2bt…(v)

Which we wish to find.Step 4: Substituting I = 5mA

16 mm ; b 15 mm ; t 2 mm

In equation ( ); we get :

3 2 25 10 16 15H 1 (mA/m)2 (16) (4 60)

or H 25.64 (mA/m)

 Ans.

46. (d)

Ampere Law states that H dl I

enclosed.

H dl = – 30 + 10 =– 20A

–30A is taken because the path taken is oppositeto the direction of magnetic field due to 30Acurrent.

47. (b)

2 2

1 x y z1

D ? Region 2 ( 1)_______ z 0D 10a 7.5a 6a

Region 1 2.5

According to boundary conditions,

1 2 1 2

2 12

2

n n t t

t tz n

2 1

x yt x y

D D E ED D

6a D

10a 7.5aD 4a 3a

2.5

2 22 t n x y zD D D 4a 3a 6a

48. (c)

AB

C

+Q

–2Q

+Q

a 30,2

a,02

a,02

0

Let the point P(0, y)Potential at point P(0, y) due to all three charges

V1 =0

1 Q KQ4 AP AP

V2 =0

1 Q KQ4 BP BP

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V3 =0

1 2Q 2KQ4 CP CP

= 1 2 3KQ KQ 2KQV V V 0AP BP CP

AP BP

or,2KQAP =

2KQCP

or, AP = CP2

2a y4

= a 3y2

or,2

2a y4

=2

2 a 3y ay 34

y a 3 =2a

2

y =a

2 3Therefore, the coordinates of the point P is

a0,P2 3

.

49. (a)The general point in z = 0 plane is P (x,y,0)

The vector directed from the line towards thispoint is

R

= x y zˆ ˆ ˆa a ay 2 0 5x x

= y zˆ ˆa 5ay 2

The direction of electric field will be same asthat of the vector.

y z

2 2

ˆ ˆa 5ay 2

y 2 5

= y z1 2ˆ ˆa a3 3

y 2

5

= 12

y = 0.5

R

= y zˆ ˆ2.5a 5a

E

= 20

R2 R

l

= 9 9

y zˆ ˆ2.5a 5a16 10 9 102 31.25

= y zˆ ˆ23a 46a

50. (a)

Taking an element current Idz

z Q

RI

xP S

I

R

d1 d2

y

Magnetic vector potential A1 due to wire PQ

A1 = L

o

-L

dzI4 R

A1 = L

o2 2

0 1

I dz2 d + z

A1 = Lo 22

1 0

zln z + z + d

2

A1 = o 2211

Iln – lndL + L + d

2

A1 = o1

Iln – lnd2L2

…(1)Similarly, for wire RS

o2 2A = – ln – lnd2L2

Resultant A = A1 + A2

A = o2 1

Ilnd – lnd

2

= 22o21

dIln

d4

51. (c)

×

M N

H2H1

Here, = 30°

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and H1 = 2I 5H

2 a 2

H1 is due to 10 Av and H2 is due to 10A at N atM.

H1 cos and H2 cos will cancel each other.

Net field intensity = H1 2sin H sin

= 2H1 sin 30° =

5 122 2

H = 5 A

m2 .

52. (c)

From the right hand rule, we can say, that

magnetic field due to current flowing is in adirection.

So, using Ampere’s law—

For 0 6 :

2 .H = 16

H = 162

For 6 10 :

2 H = 16 – 12 = 4

H = 4

2

So, H

= 16 a2 0 6

= 4 a

2 0 10

B

=

016a

2 0 6

=

04a

2 6 10

Total flux,

= 0.010.06 0.010.07

0 0z

0 0.01 0 0.06

16 4ˆ ˆˆ ˆd dza d d aB.ds a . a2 2

= 0.06 0.01 0.070.010

z00.01 0 0.06

4 d d4 d dz2

= 044(0.01) ln(6) (0.01)ln(7/6)

2

= 58.4 nWb.

53. (b)

Since, Magnetic field intensity,

H

= n1 ˆJ a2

= x z1 ˆ ˆa a30 402

= x zˆ ˆ5a a

= x z5 ˆ ˆa a

= yˆ5a A m

54. (b)

Current density J H

J

=

z

5 2

ˆ ˆ ˆa a a

1z

0 .(10 ) 0

J

= 55 3z z

1 ˆ ˆ. a 3 10 a10

The total current flowing in the wire is—

I = 2 0.0050.005 5

zz00 0

ˆJ.ds . d d aˆ3 10 a

= 3 × 1052 0.005 20 0

d d

= 3 × 105 ×3(0.005)2 0.079A

3

R = V 0.1 1.26I 0.079 .

55. (d)

Magnetic field due to a solenoid with ‘N’ turnsper unit length and current I will be:B = µ0NI

=

72

1004 10 510 10

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B = 3 b2

W2 10m

56. (d)

Given,

23 2x y z31 2F a a a3x K zK xy K z 3xz y

F =

x y z

23 231 2

a a a

x y z3x K zK xy K z 3xz y

= 2 2x y z3 12a a a1 K 6x K x3z K 3z

= 2x y z3 2 1a 3z a x aK 1 K 1 6 K

Given vector field F is irrotational i.e.

F = 0

2x y z3 2 1a 3z a x aK 1 K 1 6 K = 0

if 3 2 1K 1 0, K 1 0, 6 K = 0 3 2 1K 1, K 1, K = 6

3x1 2

2x y z y32 z

aK xy K za a a.F . a3x K zx y z a3xz y

=

23 2

31 2 3x K zK xy K z 3xz yx y z= 1K y 0 6xzAt point (1, 1, –2)

.F = 1K 61 1 2

= 112 K

= 12 6 6 1K 6

57. (d)

V = 5

x y zˆ ˆ ˆ2a 3a 4a 10 m/sec

B = x y zˆ ˆ ˆ2a 2a a mT

V B

= x y z

2

ˆ ˆ ˆa a a10 2 3 4

3 2 1

= 2x y zˆ ˆ ˆa (3 8) – a (– 2 – 12) a (4 9)10

= x y zˆ ˆ ˆ1100a 1400a 500a

Total force, F = q E V B

= 16x y z xˆ ˆ ˆ ˆ2 10 [100a 200a 300a 1100a

x y zˆ ˆ ˆ1100a 1400a 500a ]

= 16x y zˆ ˆ ˆ1200a 1200a – 200a2 10

F = 14

x y zˆ ˆ ˆ6a 6a a4 10

The acceleration,

a =

14x y z

26

ˆ ˆ ˆ6a 6a a4 10Fm 5 10

a = 8 × 1011 x y zˆ ˆ ˆ6a 6a a .

58. (d)Given, H1 = 4ax + 6ay + 8az

Therefore, H1t = 6ay + 8az

H1n = 4ax

According to magnetic boundary conditions,

1 1n 2n 2 1t 2t

o x o 2n

2n x y z 2t

H H H H2 4a HH 8a 6a 8a H

Therefore, H2 = H2n + H2t = 8az + 6ay + 8az

Therefore comparing with H2 = pax + qay +raz

p = 8, q = 6, r = 8

59. (a)

The amount of energy required or, work done doesnot depend on the path followed, but depends onthe initial and final position i.e. displacement.So,

E = W = qE.dl

= q E.d l

= 2

x y z x y zˆ ˆ ˆ ˆ ˆ ˆ10 2xa 3y a 4a dxa dya dza

=3 1 1

2

0 0 0

10 2x.dx 3y dy 4 dz

= 2 2 3 310 3 0 1 0 4 1 0

= 10 4 = –40 J

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60. (b)

The magnetic flux density at a distance ‘x’ fromthe wire, is

B = 0

yI

a2 x

B =

6

y3 10 a

x

D

A B

C

x(3,0,4)(1,0,4)

i = 6 mA

(3,0,1)(1,0,1)dx

15A

X

Z

The force on differential wire of length dx is—

dF = i dl B

= 63

x y3 106 10 ˆ ˆdxa a

x

dF = 18 × 10–19

zdx ax

Total force F =

3 3–9z1 1

dx ˆdF 18 10 ax

= 18 × 10–9 |ln(x)|13

= 18 ×10–9 ln(3) za

F = 19.8 za nN.

61. (a)

The conduction current density is given by,

cJ E

65 5

c10 ˆJ 10 cos(10 t) a

5c

10 ˆJ cos(10 t) a

The conduction current will flow radially alongthe length of capacitor.

so, The total conduction current,

c cJ dS

= 510 ˆ ˆcos(10 t)a 2 a

5c 20 cos(10 t)

5c 8 cos(10 t) A

62. (d)

The magnetic field intensity at x = 0, y = 0.5 dueto current shee4t at y = 0, will be—

H = n z y

1 1 ˆ ˆk a (8a ) (a )2 2

= xˆ– 4a

The magnetic field intensity at x = 0 y = 0.5 m.due to current sheeet at y = 1 m. will be—

2H = n z y

1 1ˆ ˆ ˆk a –4a –a2 2

2H = xˆ– 2a

The net magnetic field intensity at x = 0,y = 0.5 m. will be—

H

= 1 2H H

H

= xˆ6a

The magnetic flux density

B = 0 0 xˆH – 6 a

Force per unit length on wire is—

F = i(d B)

= –3z 0 xˆ ˆ7 10 a 6 a

= 30 yˆ42 10 a

F = –52.8 ya nN.

Consider magnitude only, F = 52.8 nN63. (c)

Magnetic field in air is given y

B

= y x0 2 2 2 2x yˆ ˆa aB

x y x y

Converting trivial to cylindrical coordinates

x = r cos , y rsin

Relation between cartesian and cylindrical co-ordinates

x r y rˆ ˆ ˆ ˆ ˆa cos a sin a , a sin a cos a

Magnetic field in cylindrical coordinate is

B

=

r0 2 2 2

ˆ ˆsin a cos ar cosBr cos sin

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r2 2 2

ˆ ˆcos a sin ar sinr cos sin

= 2

r20 r

ˆ ˆsin cos a cos aB ˆ ˆsin cos a sin a

r

2 2cos sin 1

B

= 0B a ; r 0r

B = 0

0 0

aBBH Hr

By Ampere’s Law, lHd I

H = J

[By Stoke’s theorem]

H =

r z

r z

ˆ ˆ ˆa ra a

r zH rH H

=

r z

o

ˆ ˆ ˆa ra a

r zB0 r 0r

= r zˆ ˆ ˆa ra a0 0 0

H

= 0

J = H 0

64. (a)

x

y

z

x = 2

x = –2, y = 3

(2, 0, 6)

The electric field at origin due to point charge,

1E

= 9

x y z3

2 2 20

ˆ ˆ ˆ(0 2)a 0.a (0 6)a12 10

4 (2 6 )

= 9 9

x z12 10 9 10 ˆ ˆ2a 6a

253

= 1 x zˆ ˆE 0.84a 2.52a

The electric field due to line charge,

9

x y2 2 20

3 10 ˆ ˆ(0 2)a (0 3)aE2 (2 3 )

= 9 9

x y x y18 10 3 10 ˆ ˆ ˆ ˆ2a 3a 8.31a 12.46a

13

The electric field due to sheet charge,

E3 9

x0

0.2 10 a2

= xˆ11.3a

1 2 3E E E E

x z x y xˆ ˆ ˆ ˆ ˆ0.84 a 2.52 a 8.31 a 12.46 a 11.3 a

x y zˆ ˆ ˆE 3.84 a 12.4 a 2.52 a

65. (a)

q

r

For a uniformly charged ring rotating currentof frequency ‘ ’ current is given by

I = charge×revolution freq.

= qq2 2

Magnetic moment of this equivalent current

m = 2qIA . r2

= 21 .q r2

Magnetic flux density (B) : 0IB2r

q = . 2 r is charge density (c/m)

B =

0 q

2r 2

=

0 02 r4 r 2

Given 1 2 , 1 2 , r2 = 2r1

By B is independent of r.

1

2

BB = 1 1

2 21 :1