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Nadya [2.5K]
3 years ago
11

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 10-1

3 N. What is the angle between the proton's velocity and the field
Physics
1 answer:
VARVARA [1.3K]3 years ago
8 0

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

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