1. ANSWERS
OBJECTIVE UNSOLVED LEVEL - I
1. (c) 2. (b)
3. (c) 4. (c)
5. (d) 6. (a)
7. (a) 8. (a)
9. (b) 10. (a)
11. (a) 12. (b)
13. (c) 14. (a)
15. (d)
OBJECTIVE UNSOLVED LEVEL - II
1. (b) 2. (a)
3. (d) 4. (b), (c), (d)
5. (a) 6. (c)
7. (b) 8. (a)
9. (a), (b) 10. (c)
11. (c), (d) 12. (a)
13. (b) 14. (a)
15. (d)

2. SUBJECTIVE LEVEL - I (C. B. S.E.)
1. 4 4
10 T 3.1 10 T
 
 
2. 6
4 10

3. (a) 4
1.9 10 T

 normalto the plane ofthe paper going out ofit
(b) same magnitude of B

but opposite indirection to that in(a)
4. 2
8.1 10
 N; directionofforce given byFleming’s left-hand rule
5. 5
2 10
 N; attractive force normaltoAtowards B
6. 0.96 N m
7. 4 3
5 10 T 1.6 10 T
 
   towards west.
8. (b) In a smallregionoflengh2d about the mid-point between the coils.
3 / 2 3/ 2
2 2
2
2 2
0 IR N R R
B d R d R
2 2 2
 
 
   
    
   
 
      
   
   
 
   
   
   
 
3/ 2 3/ 2 3 / 2
2 2
0 IR N 5R 4d 4d
1 1
2 4 5R 5R
  
 
      
    
 
     
   
 
   
3/ 2
2
0 IR N 4 6d 6d
1 1
2 5 5R 5R

    
    
   
   
where in the second and third steps above, terms containing 2 2
d / R and higher
powers ofd/Rare neglected since
d
1
R
 . The terms linear in d/Rcancle giving a
uniformfieldBinasmallregion:
3/ 2
0 0
IN IN
4
B 0.72
5 R R
 
 
  
 
9. (a) Cicular trajectoryofradius 1.0 mmnormalto B

(b) helicaltrajectoryofradius 0.5 mmwith velocitycomponent 7 1
2.3 10 ms
 along
B

.
10. Deuteriumions ordeuterons;the answer isnot unique because onlythe ratio ofcharge
ofmass is determined. Other possible answers are He++
, Li+++
etc.

3. 11. (a) Ahorizontalmagneticfield ofmagnitude 0.26T normalto theconductor in such
a directionthat Fleming’s left-hand rule gives amagnetic force upwards.
(b) 1.176 N.
12. (a) 2.1 N verticallydowarwards
(b) 2.1 N verticallydowards (true for anyangle between current and direction and
B since sin l  remains fixed, equalto 20 cm)
(c) 1.68 N verticallydowards.
13. Use IA B
  
 

and F I B
l
 

 
(a) 2
1.8 10 N

 malong - ydirection
(b) same as in (a)
(c) 2
1.8 10 N

 malong -x direction
(d) 2
1.8 10 N

 mat an angle of240º withthe +x direction
(e) zero
(f) zero
Force is zero in each case. Case (e) corresponds to stable, and case (f) corresponds to
unstableequilibrium.
14. (a) zero
(b) zero
(c) force on each electron is 25
evB IB/(nA) 5 10 N

   . NoteAnswer
(d) denotes onlythe magnetic force.
15. 4
7.8 10 N

 , towards the long conductor ifthe current in the closer side oftheloop is in
the same directionas the current inthe long conductor;repulsive otherwise. Not that the
forces onthe two smaller sides ofthe loop canceleach other.
SUBJECTIVE UNSOLVED LEVEL - II
1. 1
ˆ
F 1.2iN
 , 2
ˆ
F 1.2jN
 , 3
ˆ ˆ
F ( 1.2i 1.2j)N
   , 4
ˆ
F 1.2jN
 .
2.
0
mv sin
qE

3.
1 mv
2tan
qBR
  
   
 

4. 4. (a) 1.13N (b) 1.25 N
5. (a)
mv
qB (b) 2
  
(c)
m
( 2 )
qB
   (d)
mv m
, 2 ( 2 )
qB qB
      .
6. 0Qrni
2m

7. 0.115 T
8. Proton willmove incircular path ofradius 50 2 and Neutronwillmove tangentiallybe-
cause neutrondoes not have anycharge to be incircular path.
9. 4.4 Amp
10. 2
  rad
SUBJECTIVE UNSOLVED LEVEL - III
2. 0
2qE Z
m
3. (a) 0
2
R
3
  (b)
4
4
R
3

4. (a)
qBd
2m
(b)
d 3d
,
2 2
(c)
m
6qB

(d) the particlesstick together and thecombined mass moves withconstant speed m
v
along the straight line drawn upward in the plane of figure through the point of
collision.
5. 0
qB
x(t) Rcos t
m
 
  
 
 
0
qB
y(t) Rsin t
m
 
  
 
 
0
 isinitiglangular positionofthe particle.

5. 6. 2 ai Bsin
  , out ofthe page
7. 1.63 A
8.
2
3mv qB
x(t) Rsin t
qE m
 
   
 
,
2
2
mv m 1 qE
y(t) t
2qE 2 m
 
qB
z(t) R 1 cos t
m
 
 
    
 
 
 
.
9. (a)
2
0
2 2
Ia
2 c(b a )

 
(b)
0
2 2
Ic
2 (b a )

 
10. 0 0
II sin
(b a)
 


PROBLEMS
1. (a) 5
B 6.54 10 T

 
(b) Force acing onthe wire at the centre is zero
Force on arc AC = 0
force on segment CD is 6
8.1 10 N

 (inwards)
2. (i)
d
x
3
  (ii) 0
i
f
2 d


 
3.
3
min
2 mv cos
B 4.73 10 T
q(GS)
 
  
4.
2 2 2
0
2
I L a ˆ
F ln k
2 a
  

   
  

, zero.