91Ó°ÊÓ

$$\begin{gathered} v=7.20×{10}^{6}m/s \\ B=83×{10}^{-3}T \\ e=1.6×{10}^{-19}C \\ {m}_e=9.1×{10}^{-31}kg \\ a=4.90×{10}^{14}m/s^2 \end{gathered}$$

Magnetic force can be written from equation 28-3 as a vector product of velocity and magnetic field. From that, we can calculate largest value of force when velocity vector and magnetic field are perpendicular to each other. The smallest value of force can be calculated when velocity vector and magnetic field are both parallel to each other. After that, we can find the angle between the velocity vector and magnetic field by using Newton’s law.

Formula:

$F⃗=q(v⃗×B⃗)$

The largest value of force occurs if the velocity and magnetic field are perpendicular to each other.

$$\begin{gathered} F=q\left(\stackrel{→}{V}×\stackrel{→}{B}\right) \\ =qvB\sin \mathrm{θ} \end{gathered}$$

$\mathrm{θ}={90}^{∘}$

F = qvB

=$\left(1.6×{10}^{-19}C×7.20×{10}^{6}m/s×83×{10}^{-3}T\right)$

The smallest value of the magnetic force when the velocity and the magnetic field are parallel to each other:

$$\begin{gathered} F=q\left(\stackrel{→}{V}×\stackrel{→}{B}\right) \\ =qvB\sin \mathrm{θ} \end{gathered}$$

$\mathrm{θ}=0^{∘}$

$$\begin{gathered} F=qvB\sin 0 \\ F=0N \end{gathered}$$

According to Newton’s second law, F = ma

$$\begin{gathered} a⃗=F⃗/m \\ a=qvB\sin \mathrm{θ}/m \end{gathered}$$

By rearranging the equation, we can get

$\mathrm{θ}=[ma/qvB]$

$\mathrm{θ}=\frac{(9.1×{10}^{-31}kg×4.90×{10}^{14}m/s^2)}{(1.6×{10}^{-19}C×7.20×{10}^{6}m/s×83×{10}^{-3}T)}$

$\mathrm{θ}=\left[\frac{(44.59×{10}^{-17}N)}{(956.16×{10}^{-16}N)}\right]$

$\mathrm{θ}=0.{267}^{∘}$