Question

A particle with charge 12.5muC and mass 2.9*10^-5kg moves perpendicular to a 1.02T magnetic field in a circular path of radius 22.1m, How long does it take the particle to complete one orbit? 10.2ms 10.2s 23s 14.3s 8.8s

Answer 1

The magnetic force is given by:

$F = qv\times B$

Sin the movement is perpendicular to the magnetic field:

This force is equal to the rotational movement:

$F = m\frac{v^{2}}{r}$

So:

$qvBsin(90)= m\frac{v^{2}}{r}$

$v = \frac{qBr}{m}$

$v = \frac{(12.5x10^{-6}C)(1.02T)(22.1m)}{2.9x10^{-5}kg}$

We also know that the angular speed is given by:

$\omega = \frac{v}{r} = \frac{\theta }{t}$

Where v is speed, r is radius, theta is angular displacement and t is time. And if it has to complete an orbit, the displacement is 2pi. So:

$t = \frac{\theta r}{v}$

$t = \frac{2\pi (22.1m)}{9.72m/s}$

ANSWER

ANSWER is D.

I hope it helps!!