Question

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.

Short answer

The angle between the electron velocity and the magnetic field is obtained as:

$\mathbf{10}\mathbf{.}{\mathbf{1}}^{\mathbf{\circ }\mathbf{}\mathbf{}\mathbf{}}or\mathbf{}\mathbf{}\mathbf{169}\mathbf{.}{\mathbf{9}}^{\mathbf{\circ }}$

Step-by-step solution

Given information

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 $B=1.25T$

Role of the angle between the magnetic field and velocity in magnetic force

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.

Evaluating the angles

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$\mathbf{169}\mathbf{.}{\mathbf{9}}^{\mathbf{\circ }}$ .