Question

For a particle of mass m and charge q moving in a circular path in a magnetic field B, find how does its kinetic energy depend on the radius of the curvature of its path r.
(A) kinetic energy of a particle is proportional to 1/r.
(B) kinetic energy of a particle is proportional to r^2
(C) kinetic energy of a particle is proportional to 1/r^2.
(D) kinetic energy of a particle is proportional to r.

Answer 1

Answer: option D

Explanation: when an electron is placed in a uniform magnetic field, it experiences a force, this force is responsible for the circular motion of the electron.

Hence magnetic force = centripetal force.

qvB = mv²/r

Where q = magnitude of an electronic charge

v = velocity of electron.

m = mass of electron

B = strength of magnetic field.

r = radius of circular path.

Kinetic energy = mv²/2, but we don't have this in the equation above so we manipulate by dividing both sides of equation by 2.

qvB/2 = mv²/2r

To have the kinetic energy, we multiply both sides by "r"

qvB/2 ×r = mv²/2

On the right hand side is mv²/r which is the kinetic energy.

Hence kinetic energy = (qvB/2)×r

But the expression (qvB/2) is a constant

kinetic energy = (constant) × r

Hence kinetic energy is proportional to r

Answer 2

Answer:

B) K.E is proportional to r²

Explanation:

The motion of particle  in magnetic field is caused by lorentz force.

F=qvB--(1)

v= Velocity of particle

q=magnitude of charge

B=Strength of magnetic field

To move the object in circular path, centripetal force is

$F_c=\frac{mv^2}{r}--(2)$

Which must be equal to lorentz force. So equating (1) and(2)

$\frac{mv^2}{r}=qvB\\\\v=\frac{qrB}{m}--(3)$

K.E is given as:

$K.E=\frac{1}{2}mv^2$

Substituting values of v from (3) in above:

$K.E=\frac{1}{2}m(\frac{qrB}{m})^2\\\\K.E=\frac{(qrB)^2}{2m}$

Which shows K.E is directly proportional to r^2