Question

A singly ionized particle (i.e., with net charge +1e) moves with a constant speed when it enters a uniform magnetic field of strength 1.6 T. The particle enters the magnetic field such that its velocity is perpendicular to the magnetic field. In the magnetic field, the particle is observed to move in a circle with radius 2.00 cm once every 0.56 μs. What is the mass of the particle?

Answer 1

Mass,
`m = 2.28* 10^(-26)\ kg`

Step-by-step explanation:

Charge of the ionized particle is +1 e

Uniform magnetic field, B = 1.6 T

The particle enters the magnetic field such that its velocity is perpendicular to the magnetic field.

Radius of circle, r = 2 cm = 0.02 m

Time, t = 0.56 μs

We need to find the mass of the particle. We know that when object moves in magnetic field, centripetal force balances its motion. So,

`(mv^2)/(r)=qvB\sin\theta\\\\m=(Bqr)/(v)` .......(1)

v is velocity of particle

Velocity,
`v=(2\pi r)/(t)`

So, equation (1) becomes :

`m=(Bqr t)/(2\pi r)\\\\m=(Bqt)/(2\pi)\\\\m=(1.6* 1.6* 10^(-19)* 0.56* 10^(-6))/(2\pi)\\\\m=2.28* 10^(-26)\ kg`

So, the mass of the particle is
`2.28* 10^(-26)\ kg`.