Question

A proton moving at 3.90 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.20 10-13 N. What is the angle between the proton's velocity and the field

Answer 1

`\theta=40^0`

Step-by-step explanation:

The magnitude of the magnetic force is

`F_m=evB\sin\theta`

To find the angle, we make
`\sin\theta` subject of the formula

`\implies \sin\theta=(F_m)/(evB)=(7.20* 10^(-13))/(1.6* 10^(-19)* 3.90* 10^6* 1.80)`

`\implies \sin\theta=0.641025641`

`\therefore \theta=\sin^(-1)=39.8683^0\\\implies \theta\approxeq 40^0`