Moving Charges and Magnetism - Online Test

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q1. Two straight horizontal parallel wires are carrying the same current in the same direction, d is the distance between the wires. You are provided with a small freely suspended magnetic needle. At which of the following positions will the orientation of the needle be independent of the magnitude of the current in the wires

A. at a distance d/2 from any of the wires in the horizontal plane

B. at a distance d/2 from any of the wires

C. anywhere on the circumference of a vertical circle of a radius d and centre

D. at points halfway between the wires in the horizontal plane.

Answer : Option D
Explaination / Solution:

Consider two wires carrying current in the same direction as shown. The current acts inwards to the plane of the screen. The magnetic field lines are in the plane of the screen and are concentric circles. At the point midway between the wires, the field lines are directed opposite to each other. The magnetic fields due to

the two wires are directed opposite to each other. They also have the same magnitude since the wires carry currents of equal magnitude. At a distance d/2, in the horizontal plane, the net magnetic field is zero. A magnetic needle placed at this point experiences no force. The orientation of the needle becomes independent of the current in the wires

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q2. A beam of electrons at rest is accelerated by a potential V. This beam experiences a force F in a uniform magnetic field. The accelerating potential is changed to V’ and the force experienced by the electrons in the same magnetic field is 2F. The ratio V/V’ is

A. 1.0

B. 2.0

C. 1/4

D. 1/2

Answer : Option C
Explaination / Solution:

The energy of the electron after being accelerated by a potential V is

U = e V = \frac{1}{2} m v^{2}

v is the velocity of the electron.

v = \sqrt{\frac{2 e V}{m}}

The force it experiences in a magnetic field B is

F = B e v \sin θ

When accelerated by a potential V’, the velocity is

v^{'} = \sqrt{\frac{2 e V^{'}}{m}}

and the force in the same magnetic field is

2 F = B e v^{'} \sin θ

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q3. Two infinitely long wires carry currents in opposite directions. Then the field at a point P lying midway between them is

A. zero

B. half of the field due to each wire alone

C. twice the field due to each wire alone

D. square of the field due to each wire alone

Answer : Option C
Explaination / Solution:

When two wires carry currents in the same direction, the magnetic field lines at any point midway between them have the same direction. The magnitudes of the fields add up. If the current in the wires are the same, the field at the midpoint will have twice the magnitude of the field produced by each wire.

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q4. A wire of given length is first bent in one loop and the next time it is bent in three loops. If the same current is passed in both the cases, the ratio of the magnetic field induction at their centres will be

A. 1:4

B. 9:1

C. 1:9

D. 1:3

Answer : Option C
Explaination / Solution:

If the wire has length L, when it is coiled into a coil of 1 turn of radius r,

L = 2 π r

When it is coiled into a coil of 3 turns of radius r’

L = 3 \times 2 π r^{'}

2 π r = 3 \times 2 π r^{'}; r = 3 r^{'}

The magnetic field in the first case

B = \frac{μ_{0} I}{2 π r}

and in the second case,

B^{'} = \frac{μ_{0} n I}{2 π r^{'}} = \frac{3 μ_{0} I}{2 π \frac{r}{3}} = 9 \frac{μ_{0} I}{2 π r}

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q5. A proton (charge + e coul) enters a magnetic field of strength B (Tesla) with speed v, parallel to the direction of magnetic lines of force. The force on the proton is

A. zero

B. evB/2

C. evB

D. α

Answer : Option A
Explaination / Solution:

Lorentz force is given by

F = B q v \sin θ

When the proton enters the magnetic field parallel to the direction of the lines of force,

θ = 0

.Therefore F = 0

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q6. A uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointed in the same direction

A. the electron velocity will decrease in magnitude

B. the electron will turn to its right

C. the electron will turn to its left

D. the electron velocity will increase in magnitude

Answer : Option A
Explaination / Solution:

The force experienced by an electron in a combined action of magnetic and electric fields is

\vec{F} = - e (\vec{v} \times \vec{B} + \vec{E})

.Since the electron moves in the same direction of the magnetic field, it experiences no force due to the magnetic field.

- e (\vec{v} \times \vec{B}) = 0

.The electron is not deflected from its straight line path. The total force on the electron.

\vec{F} = - e (\vec{E})

.It experiences a force opposite to the direction of the electric field and to its direction of motion. The electron suffers retardation and its velocity decreases.

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q7. A wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire in the plane of the paper with center of ring being the center of the wire .Initially there is no current in the wire. What happens after a steady dc current is established in the wire?

A. The needles become parallel to each other

B. The needles become tangential to the ring of which they are a part

C. some needles become parallel to each other while others are uneffected

D. nothing happens

Answer : Option B
Explaination / Solution:

The current carrying wire has a magnetic field around it and the lines of force are in the form of concentric circles with their centers on the wire.

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q8. A wire of length L carrying current i is placed perpendicular to the magnetic induction B. The total force on the wire is

A. LB/i

B. iB/L

C. iL/B

D. iLB

Answer : Option D
Explaination / Solution:

Magnitude of the Force experienced by a current carrying conductor placed in a magnetic field is . If the angle between the directions of the current and the magnetic field is 90⁰, F=iLB

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q9. A steel wire of length l has a magnetic moment M. It is then bent into a semicircular arc. The new magnetic moment is

A. M / l

B. M

C. M×l

D. 2M/π

Answer : Option D
Explaination / Solution:

If the steel wire is assumed to be a bar magnet of length l and pole strength qm, its magnetic moment

M = q_{m} l

When it is bent into a semicircular arc of radius r, the distance between the poles is equal to its diameter

l^{'} = 2 r

The new magnetic moment

M^{'} = q_{m} l^{'} = q_{m} (2 r)

.Since,

l = π r

\frac{M^{'}}{M} = \frac{q_{m} (2 r)}{q_{m} l} = \frac{2 r}{l} = \frac{2 r}{π r} = \frac{2}{π}

M^{'} = \frac{2 M}{π}

# Moving Charges and Magnetism # CBSE 12TH PHYSICS Prepare / Learn

Q10. A wire of given length is first bent in one loop and then in three loops. If same current is passed in both cases, the ratio of magnetic inductions at the centre will be

A. 9 : 1

B. 1 : 3

C. 1 : 9

D. 1 : 4

Answer : Option C
Explaination / Solution:

1 : 9

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