Moving Charges and Magnetism Practice Questions - DPP for JEE

 Page 1


1. (d)
...(i)
When the current is reversed in I
2
,
B
2
 = – ...(ii)
Dividing (ii) by (i) we get
– (i
1
 + i
2
) = 5i
2
 – 5i
1
 ? 6i
2
 = 4i
1
2. (b) Magnetic field due to segment ‘1’
= 
Magnetic field due to segment 2
Page 2


1. (d)
...(i)
When the current is reversed in I
2
,
B
2
 = – ...(ii)
Dividing (ii) by (i) we get
– (i
1
 + i
2
) = 5i
2
 – 5i
1
 ? 6i
2
 = 4i
1
2. (b) Magnetic field due to segment ‘1’
= 
Magnetic field due to segment 2
?  at centre
 = 
3. (d) Torque on the solenoid is given by
= MB sin ?
where ? is the angle between the magnetic field and the axis of solenoid.
M = niA
  = niA B sin 30°
4. (c) Time period of cyclotron is
; ;  
? p = eBR = R = 2pm?R
K.E. = = 2p
2
m?
2
R
2
5. (a) R
g
 = 50?, I
g
 = 25 × 4 × 10
–A
? = 10
–2
 A
Range of V = 25 volts
V = I
g
(R
e
 + R
g
)
6. (b) The wires A and C carry current in same direction, therefore they
attract each other. The force on C due to is A towards the wire A
and is given by.
F
CA
 =  
Page 3


1. (d)
...(i)
When the current is reversed in I
2
,
B
2
 = – ...(ii)
Dividing (ii) by (i) we get
– (i
1
 + i
2
) = 5i
2
 – 5i
1
 ? 6i
2
 = 4i
1
2. (b) Magnetic field due to segment ‘1’
= 
Magnetic field due to segment 2
?  at centre
 = 
3. (d) Torque on the solenoid is given by
= MB sin ?
where ? is the angle between the magnetic field and the axis of solenoid.
M = niA
  = niA B sin 30°
4. (c) Time period of cyclotron is
; ;  
? p = eBR = R = 2pm?R
K.E. = = 2p
2
m?
2
R
2
5. (a) R
g
 = 50?, I
g
 = 25 × 4 × 10
–A
? = 10
–2
 A
Range of V = 25 volts
V = I
g
(R
e
 + R
g
)
6. (b) The wires A and C carry current in same direction, therefore they
attract each other. The force on C due to is A towards the wire A
and is given by.
F
CA
 =  
 0.15
or F
CA
 = 3 x 10
-5
 N (towards left).
Similarly, the wires B and C attract each other as they also carry the
currents in same direction. the force on C due to current in B is
towards right hand side. Therefore, the force on C due to B is given
by
F
BC 
= 
=
or F
BC
 = 6 × 10
–5
 N (towards right)
Therefore, the net force on C is
F = (6 × 10
–5
 – 3 × 10
–5
)
= 3 × 10
–5
 N (towards right).
7. (d) Magnetic field at the centre of the current loop is
or, 
Substituting the given values, we get
 = 1.25 m
8. (a) The magnetic field at O due to current in DA is
(directed vertically upwards)
The magnetic field at O due to current in BC is
(directed vertically downwards)
The magnetic field due to current AB and CD at O is zero.
Therefore the net magnetic field is
Page 4


1. (d)
...(i)
When the current is reversed in I
2
,
B
2
 = – ...(ii)
Dividing (ii) by (i) we get
– (i
1
 + i
2
) = 5i
2
 – 5i
1
 ? 6i
2
 = 4i
1
2. (b) Magnetic field due to segment ‘1’
= 
Magnetic field due to segment 2
?  at centre
 = 
3. (d) Torque on the solenoid is given by
= MB sin ?
where ? is the angle between the magnetic field and the axis of solenoid.
M = niA
  = niA B sin 30°
4. (c) Time period of cyclotron is
; ;  
? p = eBR = R = 2pm?R
K.E. = = 2p
2
m?
2
R
2
5. (a) R
g
 = 50?, I
g
 = 25 × 4 × 10
–A
? = 10
–2
 A
Range of V = 25 volts
V = I
g
(R
e
 + R
g
)
6. (b) The wires A and C carry current in same direction, therefore they
attract each other. The force on C due to is A towards the wire A
and is given by.
F
CA
 =  
 0.15
or F
CA
 = 3 x 10
-5
 N (towards left).
Similarly, the wires B and C attract each other as they also carry the
currents in same direction. the force on C due to current in B is
towards right hand side. Therefore, the force on C due to B is given
by
F
BC 
= 
=
or F
BC
 = 6 × 10
–5
 N (towards right)
Therefore, the net force on C is
F = (6 × 10
–5
 – 3 × 10
–5
)
= 3 × 10
–5
 N (towards right).
7. (d) Magnetic field at the centre of the current loop is
or, 
Substituting the given values, we get
 = 1.25 m
8. (a) The magnetic field at O due to current in DA is
(directed vertically upwards)
The magnetic field at O due to current in BC is
(directed vertically downwards)
The magnetic field due to current AB and CD at O is zero.
Therefore the net magnetic field is
(directed vertically upwards)
9. (c) To keep the main current in the circuit unchanged, the resistance of
the galvanometer should be equal to the net resistance.
10. (c) As electron move with constant velocity without deflection. Hence,
force due to magnetic field is equal and opposite to force due to
electric field.
qvB = qE ? v = 
11. (d) Current in a small element, 
Magnetic field due to the element
The component dB cos ?, of the field is cancelled by another opposite
component.
Therefore,
Page 5


1. (d)
...(i)
When the current is reversed in I
2
,
B
2
 = – ...(ii)
Dividing (ii) by (i) we get
– (i
1
 + i
2
) = 5i
2
 – 5i
1
 ? 6i
2
 = 4i
1
2. (b) Magnetic field due to segment ‘1’
= 
Magnetic field due to segment 2
?  at centre
 = 
3. (d) Torque on the solenoid is given by
= MB sin ?
where ? is the angle between the magnetic field and the axis of solenoid.
M = niA
  = niA B sin 30°
4. (c) Time period of cyclotron is
; ;  
? p = eBR = R = 2pm?R
K.E. = = 2p
2
m?
2
R
2
5. (a) R
g
 = 50?, I
g
 = 25 × 4 × 10
–A
? = 10
–2
 A
Range of V = 25 volts
V = I
g
(R
e
 + R
g
)
6. (b) The wires A and C carry current in same direction, therefore they
attract each other. The force on C due to is A towards the wire A
and is given by.
F
CA
 =  
 0.15
or F
CA
 = 3 x 10
-5
 N (towards left).
Similarly, the wires B and C attract each other as they also carry the
currents in same direction. the force on C due to current in B is
towards right hand side. Therefore, the force on C due to B is given
by
F
BC 
= 
=
or F
BC
 = 6 × 10
–5
 N (towards right)
Therefore, the net force on C is
F = (6 × 10
–5
 – 3 × 10
–5
)
= 3 × 10
–5
 N (towards right).
7. (d) Magnetic field at the centre of the current loop is
or, 
Substituting the given values, we get
 = 1.25 m
8. (a) The magnetic field at O due to current in DA is
(directed vertically upwards)
The magnetic field at O due to current in BC is
(directed vertically downwards)
The magnetic field due to current AB and CD at O is zero.
Therefore the net magnetic field is
(directed vertically upwards)
9. (c) To keep the main current in the circuit unchanged, the resistance of
the galvanometer should be equal to the net resistance.
10. (c) As electron move with constant velocity without deflection. Hence,
force due to magnetic field is equal and opposite to force due to
electric field.
qvB = qE ? v = 
11. (d) Current in a small element, 
Magnetic field due to the element
The component dB cos ?, of the field is cancelled by another opposite
component.
Therefore,
12. (a) The direction of  is along 
? The magnetic  force
 is along OY.
13. (a) ;  
In ? CBD,  
14. (a) Let us consider a thickness dx of wire. Let it be at a distance x