Question

A particle of mass 2.4×10^−6 kg and charge 27 μC is moving at a speed of 27 m/s perpendicularly to a uniform 9.5-T magnetic field. What is the radius of the particle’s circular path?

Answer 1

0.253 m

Step-by-step explanation:

Given that, the mass of the particle,
`m=2.4*10^(-6) kg`.

Charge of the particle,
`q=27*10^(-6)  kg`.

Speed of the particle,
`v=27 m/s`.

Magnetic field is,
`B=9.5 T`.

The radius of the particle's circular path when the speed of the particle is perpendicular to the magnetic field is,

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

By putting all the values in the above equation.

`r=((2.4*10^(-6)kg)(27m/s) )/((27*10^(-6)C)(9.5T) )`.

Therefore by further solving,

`r=0.253m`

Therefore the radius of the particle in a circular path is 0.253 m