Question

The force on a charged proton (charge 1.6 E-19) travelling at 3 E6 m/s through a magnetic field of 4 T is:
(note: E represents power of ten)

Answer 1

Answer on Question #77351, Physics / Electromagnetism

Question. The force on a charged proton (charge $1.6 \cdot 10^{-19} C$) travelling at $3 \cdot 10^{6} \, \text{m/s}$ through a magnetic field of $4 \, \text{T}$ is:

Solution.

According to the Lorentz force

$$\vec{F} = q \vec{E} + q [\vec{v}, \vec{B}]$$

So, $\vec{E} = 0$. We get

$$F = q v B \cdot \sin \alpha$$

If $\alpha = 90{}^{\circ}$ then $\sin 90{}^{\circ} = 1$. Thus

$$F = q v B \cdot \sin \alpha = 1.6 \cdot 10^{-19} \cdot 3 \cdot 10^{6} \cdot 4 \cdot 1 = 19.2 \cdot 10^{-13} \, \text{N}$$

If $\alpha = 0{}^{\circ}$ then $\sin 0{}^{\circ} = 0$. Thus

$$F = 0$$

If $0 < \alpha < 90{}^{\circ}$ then

$$F = q v B \cdot \sin \alpha = 1.6 \cdot 10^{-19} \cdot 3 \cdot 10^{6} \cdot 4 \cdot \sin \alpha = 19.2 \cdot 10^{-13} \cdot \sin \alpha, \, \text{N}$$

Answer. $F = 19.2 \cdot 10^{-13} \cdot \sin \alpha, \, \text{N}$. If $\alpha = 0{}^{\circ}$, $F = 0$; if $\alpha = 90{}^{\circ}$, $F = 19.2 \cdot 10^{-13} \, \text{N}$.

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