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

Answer:

$\theta=40^0$

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

$\implies \sin\theta=\frac{F_m}{evB}=\frac{7.20\times 10^{-13}}{1.6\times 10^{-19}\times 3.90\times 10^6\times 1.80}$

$\implies \sin\theta=0.641025641$

$\therefore \theta=\sin^{-1}=39.8683^0\\\implies \theta\approxeq 40^0$