Question

A charge partides round a 1 m radius circular particle accelerator at nearly the speed of light. Find : (a) The period (b) The centripetal acceleration of the charged particles

Answer 1

Explanation:

It is given that,

Radius of circular particle accelerator, r = 1 m

The distance covered by the particle is equal to the circumference of the circular path, d = 2πr

d = 2π × 1 m

(a) The speed of satellite is given by total distance divided by total time taken as :

$speed=\dfrac{distance}{time}$

Let t is the period of the particle.

$t=\dfrac{d}{s}$

d = distance covered

s = speed of particle

It is given that the charged particle is moving nearly with the speed of light

$t=\dfrac{d}{c}$

$t=\dfrac{2\pi\times 1\ m}{3\times 10^8\ m/s}$

$t=2.09\times 10^{-8}\ s$

(b) On the circular path, the centripetal acceleration is given by :

$a=\dfrac{c^2}{r}$

$a=\dfrac{(3\times 10^8\ m/s)^2}{1\ m}$

$a=9\times 10^{16}\ m/s^2$

Hence, this is the required solution.