SOLUTIONS

1. (b) : Net force on – q towards the centre.
   F = (2F 1 ·sinq) =
   
   
   • 
     
   
   
   
   

×

+

(^222)
22

.

KQq

dR

R

dR

For particle to move in a circle

F = mZ^2 R

Ÿ 222 KQqR 32 2

dR

mR

()+ /

=ω

Ÿ ω=

+

2

2232

KQq

md()R /

=

×× ×××

××+

=

−

−

2910 10 10 10 −

90 10 34

400

96

32232

1

()/

rads

2. (a) : Work done for moving a charge q 0 from B to A.
   WB o A = q 0 (VA – VB)
   Also, VA = potential at point A

=+=+









kq

r

kq

r

k

q

r

q

r

1

1

2

2

1

1

2

2

=

−×

×

+

×

×

















=

−

−

−

(^10) −

510

15 10

210

510

10

15

10 6

2

6

2

6

V

Similarly,

VB=

×

×

−

×

×

















=

−

×

−

−

−

(^10) −

210

15 10

510

510

13

15

10 6

2

6

2 10 V

6

⇒=××−

−

×





















WBA→ 310 − =

1

15

10

13

15

(^6610628) .J

3. (b) : When S 1 is closed and S 2 is open, w 1 is in circuit.
   Current through w 1.

I

Rr

p

1 = +

ε

()

Ÿ (^) I 1 pp
213

=

+

=

εε

Potential dierence across l

2

length of w 1

==

1

2

2

(^13)

[]I

εp

is should be equal to H So,

εp

3

= H Ÿ Hp = 3H

Similarly for second case : I

RR

p

(^21)

3

1

=

+

=

+

ε ε

Potential dierence across 2l/3 length of w 2

==

+









=

+

2

3

2

3

3

1

2

(^21)

IR

R

R

R

R

εε

is should be equal to H. So,

2

1

ε

ε

R

R+

= Ÿ R = 1 :

4. (c) : Number density of electrons,

n=

×

=







=





− 

(^610) −−
63 9
6
7
10
6
7
10
23
3
23 3293
()gg/( cm )
cm m
B
Q Q
A d
R
F 1 F 1
d
–q
A = pr^2 = p(0.5 × 10–3m)^2 = 0.25p × 10–6 m^2
v

I

d neA

=

=

−−××−

11

671029 3116 10 9602510 2

.

( /) (. )

A

mCπ m

= 0.1 mm s–1

5. (b) : As I

VV

ABC= 44 ΩΩ+ = 8 Ω,

()VVAB−=VV









=

8

4

Ω 2

Ω

Similarly, (VA – VD) =

V

4

us, (VD – VB) = (VD – VA) + (VA – VB)

=

−

+=

VVV

424

Since, (VD – VB) > 0, current ows from D to B.

6. (a) :



Fqe= E

=−()16 10×−−^18 () 104 k

=×16 10−^14 k

  

Fqm=×()vB

=−()16 10××−^18 () 10 iBj

=−()16 10× −^17 Bk

From gure, since

 

FFme+=0,FFme=−

or 16 × 10–17 B = 16 × 10–14

Hence, B = 10^3 Wb m–2

7. (a) : Refer to gure. B I
   AB OC

=−−

μ

π

(^0) αα
4( )
[sin sin( )]

μ
πα
(^0) α
4

2

I

Rcos

(sin ) (as OC = R cos D)

or B

I

AB R

=







⊗

μ

π

(^0) α
2
tan

B

I

ADB R

=









−













⊗

μ

π

(^0) πα
4

() 22

=−













⊗

μ

π

(^0) πα
2

I

R

()

us, B = BADB + BAB =−+

μ

π

(^0) πα α
2

I

R

(tan )

8. (a) : Angular frequency of oscillation
   ω= =
   ××−−×

11

LC 1 02. 1036810

= 2.5 × 10^4 rad s–1 = 25 × 10^3 rad s–1

Q

C

L

dI

dt

Q

C

L

dQ

11 dt

2

−= 00 ⇒+ 2 =

+Z

–Z

Fe

v

Bo Fm

Eo

o

o

o

q X

Y

(2p – 2D)

R

D

A C B

O

D D

I