Acharge 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

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 :

Let t is the period of the particle.

d = distance covered

s = speed of particle

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

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

Hence, this is the required solution.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach