Question

A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction. the mass of a proton is 1.673 × 10−27 kg. find the magnitude of the field. answer in units of t.

Answer 1

The magnetic force experienced by the proton is given by

`F=qvB \sin \theta`
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
`\theta` the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so
`\sin \theta=1` and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:

`F=ma`

So we have

`ma=qvB`
from which we can find the magnitude of the field:

`B= (ma)/(qv)= ((1.67 \cdot 10^(-27)kg)(5\cdot 10^(13)m/s^2))/((1-6 \cdot 10^(-19)C)(3.7 \cdot 10^7 m/s))=0.014 T`