# Moving Charges And Magnetism

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| By Tanmay Shankar
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Tanmay Shankar
Community Contributor
Quizzes Created: 547 | Total Attempts: 1,821,568
Questions: 15 | Attempts: 1,827  Settings  Time: 30 Minute

• 1.

### A wire is placed parallel to the lines of force in mangnetic field and current flows in the wire. Then

• A.

The wire will experience a torque

• B.

The wire will experience a force along the direction of magnetic field

• C.

The wire will not experience any force

• D.

The wire will experience a perpendicular force.

C. The wire will not experience any force
Explanation
When a wire is placed parallel to the lines of force in a magnetic field and current flows through it, the wire will not experience any force. This is because the magnetic force on a current-carrying wire is given by the equation F = BILsinθ, where B is the magnetic field strength, I is the current, L is the length of the wire, and θ is the angle between the wire and the magnetic field. When the wire is parallel to the magnetic field lines, the angle θ is 0, resulting in sinθ = 0. Therefore, the force on the wire is zero and it does not experience any force.

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• 2.

### A circular coil of radius 4 cm and 20 turns carries a current of 3A. It is placed in a magnetic field of induction 0.5T. The magnetic dipole moment of the coil is

• A.

0.6Am2

• B.

0.45 Am2

• C.

0.3 Am2

• D.

0.15 Am2

C. 0.3 Am2
Explanation
The magnetic dipole moment of a coil is given by the formula M = nIA, where n is the number of turns, I is the current, and A is the area of the coil. In this case, the number of turns is 20, the current is 3A, and the radius of the coil is 4 cm. The area of the coil can be calculated using the formula A = πr^2. Plugging in the values, we get A = π(0.04m)^2 = 0.005 m^2. Substituting all the values into the formula for the magnetic dipole moment, we get M = 20 * 3A * 0.005 m^2 = 0.3 Am^2. Therefore, the correct answer is 0.3 Am2.

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• 3.

### Currents of 10 ampere and 2 ampere are passed through two parallel wires A and B respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is 2m. The force acting on the conductor B, which is situated at 10 cm distance form A will be

• A.

8 X 10–5N

• B.

5 X 10–5N

• C.

8π X 10–7N

• D.

4π X 10–7N

A. 8 X 10–5N
Explanation
When two parallel wires carry currents in opposite directions, they create a magnetic field that interacts with each other. The force between the wires can be calculated using the formula F = (μ₀ * I₁ * I₂ * L) / (2 * π * r), where F is the force, μ₀ is the permeability of free space, I₁ and I₂ are the currents in the wires, L is the length of wire B, and r is the distance between the wires. In this case, since wire A is infinitely long, we can assume that the force acting on wire B is solely due to the magnetic field created by wire A. Therefore, the force acting on wire B will be 8 X 10⁻⁵N.

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• 4.

### To convert a galvanometer into an ammeter, we connect

• A.

Low resistance in series

• B.

Low resistance in parallel

• C.

High resistance in series

• D.

High resistance in parallel

B. Low resistance in parallel
Explanation
To convert a galvanometer into an ammeter, we connect a low resistance in parallel. This is because an ammeter is used to measure current, and it needs to have a very low resistance in order to minimize the voltage drop across it and ensure accurate current measurement. By connecting a low resistance in parallel with the galvanometer, we provide an alternative path for current to flow, bypassing most of it away from the galvanometer coil. This allows the majority of the current to pass through the low resistance, resulting in a small voltage drop and accurate current measurement on the ammeter.

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• 5.

### Two parallel conductors are carrying current 'i' in the same direction. Force experienced per unit length by each conductor (distance between the conductors is r) would be

• A.

Repulsion and

• B.

Repulsion and

• C.

Attraction and

• D.

Attraction and

D. Attraction and
Explanation
When two parallel conductors are carrying current in the same direction, they create magnetic fields around them that interact with each other. According to the right-hand rule, the magnetic field lines produced by one conductor will be in the opposite direction to the magnetic field lines produced by the other conductor. This results in an attraction force between the two conductors. Therefore, the correct answer is "Attraction and".

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• 6.

### If a particle is moving in a uniform magnetic field, then

• A.

Its momentum changes but total energy ramains the same

• B.

Both momentum and total energy remains the same

• C.

Its total energy changes but momentum remains same

• D.

Both momentum and tatal energy will change

A. Its momentum changes but total energy ramains the same
Explanation
When a particle is moving in a uniform magnetic field, its momentum changes due to the Lorentz force acting on it. However, the total energy of the particle remains the same. This is because the magnetic force only affects the particle's motion perpendicular to the magnetic field, causing it to move in a circular path. As a result, the particle's kinetic energy may change, but its total energy (which includes both kinetic and potential energy) remains constant.

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• 7.

### A particle A has charge + q and a particle B has charge + 4q with each of them having the same mass m. When allowed to fall from rest through the same electric potential difference, the ratio of their speeds uA / uB will therefore:

• A.

2 : 1

• B.

1 : 2

• C.

1 : 4

• D.

4 : 1

B. 1 : 2
Explanation
When particles A and B are allowed to fall through the same electric potential difference, the ratio of their speeds can be determined using the equation for the electric potential energy gained by a charged particle. The electric potential energy gained by particle A is qV, where V is the electric potential difference, and the electric potential energy gained by particle B is 4qV. Since both particles have the same mass, the kinetic energy gained by each particle is equal to their respective electric potential energies. Therefore, the ratio of their speeds is equal to the square root of the ratio of their electric potential energies, which is equal to the square root of (qV) / (4qV), or 1/2. Thus, the ratio of their speeds is 1 : 2.

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• 8.

### E, F and G are three current carrying linear conductors. The direction of net magnetic force on F will be

• A.

In the direction of current flowing in F

• B.

In the plane of paper towards left

• C.

In the plane of paper towards right

• D.

Normal to the plane of paper

B. In the plane of paper towards left
Explanation
The direction of the net magnetic force on F will be in the plane of the paper towards the left because when current flows through a conductor, a magnetic field is created around it. According to the right-hand rule, the direction of the magnetic field around a current-carrying conductor is in a circular pattern perpendicular to the direction of the current. Since F is a current-carrying conductor, the magnetic field created by it will be in the plane of the paper. The force on F will be perpendicular to both the magnetic field and the current direction, resulting in a net force towards the left.

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• 9.

### An electron is moving along the positive X-axis. You want to apply a magnetic field for a short times so that the electron may reverse its direction and move parallel to the negative X-axis. This can be done by applying the magnetic field along

• A.

X-axis only

• B.

Y-axis only

• C.

Z-axis only

• D.

Y and Z axes

D. Y and Z axes
Explanation
To reverse the direction of the electron and make it move parallel to the negative X-axis, a magnetic field needs to be applied perpendicular to the current direction of the electron. Since the electron is moving along the positive X-axis, the magnetic field should be applied along the Y and Z axes, which are perpendicular to the X-axis. This will exert a force on the electron in a direction opposite to its velocity, causing it to reverse its direction and move parallel to the negative X-axis.

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• 10.

### A proton moving with a constant velocity passes through a region of space without change in its velocity. If E and B represent electric and magnetic fields respectively, this region of space may not have

• A.

E ≠ 0, B = 0

• B.

E ≠ 0, B ≠ 0

• C.

E = 0, B ≠ 0

• D.

E = 0, B = 0

A. E ≠ 0, B = 0
Explanation
In this scenario, the proton is moving with a constant velocity, which means that it is not experiencing any acceleration. According to the Lorentz force equation, the force on a charged particle is given by F = q(E + v x B), where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field. Since the proton is not experiencing any acceleration, the net force on it must be zero. This implies that either the electric field (E) or the magnetic field (B) must be zero. Since the proton is not experiencing any change in velocity, it cannot be influenced by an electric field (E ≠ 0), but it can still pass through a region with zero magnetic field (B = 0). Therefore, the correct answer is E ≠ 0, B = 0.

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• 11.

### Two identical magnetic dipoles of magnetic moment 1.0 Am2 each, placed at a separation of 2 m with their axes perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is

B.
Explanation
The resultant magnetic field at a point midway between the dipoles is zero. This is because the magnetic fields produced by the two dipoles cancel each other out due to their perpendicular alignment. The magnetic field produced by one dipole will have equal magnitude but opposite direction to the magnetic field produced by the other dipole. As a result, the net magnetic field at the midpoint will be zero.

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• 12.

### When a charged particle enters in a uniform magnetic field, its kinetic energy

• A.

Remains constant

• B.

Increases

• C.

Decreases

• D.

Becomes zero

A. Remains constant
Explanation
When a charged particle enters a uniform magnetic field, its kinetic energy remains constant. This is because the magnetic field does not exert any force on the charged particle in the direction of its motion. As a result, the particle's speed and hence its kinetic energy remain unchanged. The magnetic field only acts perpendicular to the particle's velocity, causing it to move in a circular path but not affecting its kinetic energy.

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• 13.

### A galvanometer of resistance 100 Ω gives full scale deflection for 20 mV. Find the resistance to be attached so that it gives full scale deflection of 5V.

• A.

in parallel

• B.

in Series

• C.

in parallel

• D.

in Series

D. in Series
Explanation
To achieve a full scale deflection of 5V on the galvanometer, a resistance needs to be attached in series with it. This is because when resistances are connected in series, the total resistance increases. By increasing the total resistance, the current passing through the galvanometer decreases, resulting in a larger voltage drop across it. Therefore, connecting the resistance in series will allow the galvanometer to give a full scale deflection of 5V.

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• 14.

### Two insulating rings, one of slightly smaller diameter than the other are suspended along their common diameter as shown. Initially the planes of the rings are mutually perpendicular. When a steady current is set up in each of them

• A.

The two rings rotate into a common plane

• B.

The inner ring oscillates about its initial position

• C.

The inner ring stays stationary while the outer one moves into the plane of the inner ring

• D.

The outer ring stays stationary while the inner one moves into the plane of the outer ring

A. The two rings rotate into a common plane
Explanation
When a steady current is set up in each of the rings, a magnetic field is produced around each ring. These magnetic fields interact with each other, causing a torque on the rings. The torque causes the rings to rotate, aligning themselves into a common plane. This is because the interaction between the magnetic fields tends to minimize the energy of the system, and aligning the rings into a common plane achieves this. Therefore, the correct answer is that the two rings rotate into a common plane.

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• 15.

### A bar magnet has a magnetic moment , and is placed in a magnetic field of 0.2 T. work done in turning the magnet from parallel to anti parallel position relative to the field direction is

• A.

0.5 J

• B.

1 J

• C.

2.0 J

• D.

Zero

B. 1 J
Explanation
When a bar magnet is turned from parallel to antiparallel position relative to the magnetic field, work is done against the magnetic field. The work done is given by the formula W = -magnetic moment x change in angle, where the negative sign indicates work done against the field. In this case, the change in angle is 180 degrees, and the magnetic moment is given. Plugging in the values, we get W = -magnetic moment x change in angle = -magnetic moment x 180 degrees. Since the magnetic moment is positive and the change in angle is 180 degrees, the work done is negative. Therefore, the correct answer is Zero.

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