Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 2

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Question 1
4 / -1

An electron with a speed of 1.8 x 10⁶ m/s is moving in a circular orbit in a uniform magnetic field of 10^-4 Wb/m², the radius of the circular path of the electron is

A
10.63 m

B
1.063 m

C
106.3 m

D
0.1063 m
Question 2
4 / -1

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

A
The electron will turn to its right

B
The electron will turn to its left

C
The electron velocity will increase in magnitude

D
The electron velocity will decrease in magnitude

Solution

As the electron is moving along the direction of the magnetic field, it will experience no magnetic force, but due to an electric force acting on it opposite to the direction of electric field (as it is a negatively charged particle) the velocity of the electron will decreases.
Question 3
4 / -1

The force acting on a charge q moving with velocityin a magnetic field is given by

A

B

C

D

Solution

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule . The force is given by the charge times the vector product of velocity and magnetic field.
Question 4
4 / -1

A Charge is fired through a magnetic field. The magnetic force acting on it is maximum when the angle between the direction of motion and magnetic field is

A
π

B
zero

C
π/2

D
π/4

Solution

The force will have a magnitude F=qvB sin q, thus it will be maximum if sin q is maximum. Thus, angle between velocity and magnetic field should be 90^o or the charge particle moves perpendicular to the velocity vector.
Question 5
4 / -1

The frequency (v) of charged particle, moving at right angles to the magnetic field is independent of

A
radius of circular trajectory (r)

B
magnetic induction (B)

C
speed of the particle (v)

D
both speed of the particle (v) and radius of trajectory (r).

Solution

Frequency (v) = qB/2πm.
Frequency is independent of radius of trajectory of particle and speed of particle.
Question 6
4 / -1

Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.

A
14

B
28

C
7

D
32

Solution

The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 * 4) = 28 units.
Question 7
4 / -1

A proton enters a magnetic field with a velocity of 1.5 x 10⁵ m/s making an angle of 30° with the field. What is the force experienced by the proton if B = 3.0 T.

A
3.6×10^-14 N

B
1.2x^-14 N

C
4.8x^-14 N

D
2.4x^-14 N
Question 8
4 / -1

When a charged particle moves in a magnetic field, its kinetic energy always

A
remain constant

B
first increases then decreases.

C
decreases

D
increases

Solution

When a charged particle enters a magnetic field B its kinetic energy remains constant as the force exerted on the particle is:

is perpendicular to so the work done by =0. This does not cause any change in kinetic energy.
Question 9
4 / -1

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in the uniform magnetic field. What should be the energy of an alpha particle to describe a circle of the same radius in the same field.

A
2 MeV

B
0.5 MeV

C
1 MeV

D
4 MeV

Solution

Kinetic energy of a charged particle,

Radius of the circular path of a charged particle uniform magnetic field is given by
Mass of a proton, m_p = m
Mass of an α-particle. m_α = 4m
Charge of a proton, q_p = e
Charge of an α -particle, q_α = 2_e
Question 10
4 / -1

An electron is moving in a circular orbit in a magnetic field 2 x 10^-3 Wb/m², the time period of electron is

A
17.85 x 10^-3sec

B
17.85 x 10^-8sec

C
17.85 x 10^-9sec

D
17.85 x 10^-5sec