Question

A charged particle enters a uniform magnetic field B_vec with a velocity v_vec at right angles to the field.It moves in a circle with period T.1. If a second particle, with the same electric charge but tentimes as massive, enters the field with the same velocity v_vec, what is its period?A. T/10B. TC. 5TD. 10T

Answer 1

Answer:

D. 10 T

Explanation:

When a particle is moving in a magnetic field, the magnetic force provides the centripetal force that keeps the particle in circular motion.

The cyclotron period (the period the particle takes to complete one orbit) can be found to be

$T=\frac{2\pi m}{qB}$

where

m is the mass of the particle

q is its charge

B is the magnetic field

As we see, the period is directly proportional to the mass of the particle.

In this problem, the second particle is ten times as massive as the first one:

m' = 10 m

while the speed is the same. So, the period of the second particle is

$T'=\frac{2\pi m'}{qB}=\frac{2\pi (10m)}{qB}=10 \frac{2\pi m}{qB}=10 T$