Q: The magnetic force on a charge q moving with velocity v at an angle \theta to a field B is:

qvB\sin\theta. The force is greatest when the velocity is perpendicular to the field and vanishes when the two are parallel.

Q: The work done by the magnetic force on a moving charged particle is:

Always zero. \vec{F} = q(\vec{v} \times \vec{B}) is always perpendicular to \vec{v}, so \vec{F}\cdot d\vec{s} = 0 and the speed never changes.

Q: The total Lorentz force on a charge moving in both electric and magnetic fields is:

q(\vec{E} + \vec{v} \times \vec{B}). The electric part acts along \vec{E} whatever the velocity; the magnetic part acts perpendicular to both \vec{v} and \vec{B}.

Question 1

The magnetic force on a charge q moving with velocity v at an angle 	heta to a field B is:

Accepted Answer

qvB\sin	heta. The force is greatest when the velocity is perpendicular to the field and vanishes when the two are parallel.

Question 2

The work done by the magnetic force on a moving charged particle is:

Accepted Answer

Always zero. \vec{F} = q(\vec{v} 	imes \vec{B}) is always perpendicular to \vec{v}, so \vec{F}\cdot d\vec{s} = 0 and the speed never changes.

Question 3

A stationary charge placed in a uniform magnetic field experiences:

Accepted Answer

No force. The magnetic force is proportional to velocity, so a charge at rest feels nothing from a magnetic field.

Question 4

A charged particle entering a uniform magnetic field perpendicular to its velocity follows:

Accepted Answer

A circular path. The constant-magnitude force always perpendicular to the velocity supplies exactly the centripetal force for uniform circular motion.

Question 5

The radius of the circular path of a charge q of mass m moving with speed v perpendicular to a field B is:

Accepted Answer

\dfrac{mv}{qB}. Equating qvB = \dfrac{mv^2}{r} gives r = \dfrac{mv}{qB}, so heavier or faster particles curve less.

Question 6

The cyclotron frequency of a charged particle in a magnetic field B is:

Accepted Answer

\dfrac{qB}{2\pi m}. Remarkably, this frequency does not depend on the speed or radius, which is what makes the cyclotron work.

Question 7

The total Lorentz force on a charge moving in both electric and magnetic fields is:

Accepted Answer

q(\vec{E} + \vec{v} 	imes \vec{B}). The electric part acts along \vec{E} whatever the velocity; the magnetic part acts perpendicular to both \vec{v} and \vec{B}.

Question 8

A charged particle entering a magnetic field at an angle other than 0^\circ or 90^\circ follows:

Accepted Answer

A helical path. The velocity component along \vec{B} is unaffected while the perpendicular component circles, so the two combine into a helix.

Question 9

An electron and a proton having the same kinetic energy enter a uniform magnetic field perpendicularly. The ratio of the radii of their paths r_e : r_p is:

Accepted Answer

\sqrt{m_e/m_p}. r = \dfrac{\sqrt{2mK}}{qB}. With K and |q| equal, r \propto \sqrt{m}, so the lighter electron has the smaller radius.

Question 10

The radius of the circular path of a charged particle of kinetic energy K in a field B is:

Accepted Answer

\dfrac{\sqrt{2mK}}{qB}. Substituting v = \sqrt{\dfrac{2K}{m}} into r = \dfrac{mv}{qB} gives this result.

Question 11

A charged particle moves in a circle in a uniform magnetic field. Its kinetic energy is proportional to the radius as:

Accepted Answer

r^2. Since r = \dfrac{mv}{qB}, v \propto r, and kinetic energy \propto v^2 \propto r^2.

Question 12

An electron moving at 10^6	ext{ m/s} enters a field of 0.1	ext{ T} perpendicularly. The radius of its path is approximately (m_e = 9.1 	imes 10^{-31}	ext{ kg}):

Accepted Answer

5.7 	imes 10^{-5}	ext{ m}. r = \dfrac{mv}{qB} = \dfrac{9.1 	imes 10^{-31} 	imes 10^6}{1.6 	imes 10^{-19} 	imes 0.1} \approx 5.7 	imes 10^{-5}	ext{ m}.

Question 13

A charge of 1.6 	imes 10^{-19}	ext{ C} moves at 10^7	ext{ m/s} perpendicular to a field of 0.5	ext{ T}. The force on it is:

Accepted Answer

8 	imes 10^{-13}	ext{ N}. F = qvB = 1.6 	imes 10^{-19} 	imes 10^7 	imes 0.5 = 8 	imes 10^{-13}	ext{ N}.

Question 14

A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is:

Accepted Answer

Zero. F = qvB\sin 0^\circ = 0, so the particle travels undeflected in a straight line.

Question 15

The time period of revolution of a charged particle in a uniform magnetic field is:

Accepted Answer

\dfrac{2\pi m}{qB}, independent of its speed. A faster particle travels a proportionately larger circle, so the time to go round is unchanged.

Question 16

A proton moving at 10^6	ext{ m/s} enters a field of 1	ext{ T} perpendicularly. The radius of its path is approximately (m_p = 1.67 	imes 10^{-27}	ext{ kg}):

Accepted Answer

1.04 	imes 10^{-2}	ext{ m}. r = \dfrac{1.67 	imes 10^{-27} 	imes 10^6}{1.6 	imes 10^{-19} 	imes 1} \approx 1.04 	imes 10^{-2}	ext{ m}.

Question 17

A charge of 2\,\mu	ext{C} moves at 10^3	ext{ m/s} making 30^\circ with a field of 0.4	ext{ T}. The force on it is:

Accepted Answer

4 	imes 10^{-4}	ext{ N}. F = qvB\sin 30^\circ = 2 	imes 10^{-6} 	imes 10^3 	imes 0.4 	imes 0.5 = 4 	imes 10^{-4}	ext{ N}.

Question 18

The cyclotron frequency of a proton in a field of 1	ext{ T} is approximately (m_p = 1.67 	imes 10^{-27}	ext{ kg}):

Accepted Answer

1.5 	imes 10^7	ext{ Hz}. f = \dfrac{qB}{2\pi m} = \dfrac{1.6 	imes 10^{-19}}{2\pi 	imes 1.67 	imes 10^{-27}} \approx 1.5 	imes 10^7	ext{ Hz}.

Question 19

If the speed of a charged particle in a magnetic field is doubled, the radius of its circular path:

Accepted Answer

Doubles. r = \dfrac{mv}{qB} \propto v, so doubling the speed doubles the radius.

Question 20

If the magnetic field is doubled, the radius of the circular path of a charged particle:

Accepted Answer

Halves. r \propto \dfrac{1}{B}, so a stronger field bends the particle into a tighter circle.

Magnetic Force & Motion of Charges — MH-CET Physics MCQs with Solutions

Questions with solutions