Question

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with period T. If a second particle, with the same electric charge but ten timesas massive, enters the field with the same velocity v, what is its period?

Answer 1

10T

Step-by-step explanation:

The radius (r) of a charged particle that moves in a circle in a magnetic field is:

`r = (mv)/(qB)` (1)

where m: is the particle's mass, v: is the speed of the particle, q: is the charge of the particle and B: is the strength of the magnetic field.

The period (T) of the charged particle is related to its radius by the following equation:

`T = (2 \pi r)/(v)` (2)

By entering equation (1) into (2) we have:

`T = (2 \pi)/(v) \cdot (mv)/(qB) = (2\pi m)/(qB) = 10 \cdto (2\pi m)/(qB) = 10T` (3)

Therefore, from equation (3) we have that if the mass increases by a ten factor, the period will increase in the same factor since they are proportional to each other.

I hope it helps you!