An electron moving at $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}\mathbf{}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$in a $\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{‐}\mathit{T}$magnetic field experiences a magnetic force of$\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{16}}\mathbf{}\mathbf{}\mathit{N}$. What angle does the velocity of the electron make with the magnetic field? There are two answers.

The speed of the electron is $v=4.00\times {10}^{3}m/s$

The magnetic force is $F=1.40\times {10}^{‐16}N$

The charge on the electron is $q=1.6\times {10}^{19}C$

The magnetic field strength is $B=1.25T$

The equation that is used to determine the angle between the electron velocity and the magnetic field is:

$\mathit{\theta }\mathbf{)}\mathbf{=}\mathit{q}\mathit{v}\mathit{B}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}$…………….(1)

where the value ofdetermines the strength of the magnetic force.

Substituting the given values in equation (1) we will get,

$\mathit{\theta }\mathit{i}\mathit{n}\mathbf{=}\mathbf{(}\mathbf{}\mathbf{}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{16}\mathbf{}\mathbf{}}\mathbf{N}}{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{19}}\mathbf{}\mathbf{}\mathbf{C}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}\mathbf{}\mathbf{}}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{T}}$

=0.175

This can give us the value of such that,

$$\begin{gathered} \mathit{\theta }\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{1}}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{175}\mathbf{)} \\ \mathbf{=}\mathbf{10}\mathbf{.}{\mathbf{1}}^{\mathbf{\circ }} \end{gathered}$$

The other value of $\theta$will also be the same as role="math" localid="1653741005917" $\mathit{\theta }\mathbf{80}\mathbf{-}$ , which will give another value of angle such that,

role="math" localid="1653741026619" $\mathit{\theta }\mathbf{801}\mathbf{‐}\mathbf{69}\mathbf{.}\mathbf{9}$

Therefore, The angle between the electron velocity and the magnetic field is obtained as,$\mathbf{10}\mathbf{.}{\mathbf{1}}^{\mathbf{\circ }}$ or$\mathbf{169}\mathbf{.}{\mathbf{9}}^{\mathbf{\circ }}$ .