An electron moves at $\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$through a regionin which there is a magnetic field of unspecified direction and magnitude $\mathbf{7}\mathbf{.}\mathbf{40}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{T}$. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle
between the electron velocity and the magnetic field?

The magnetic field force is given by

$$\begin{gathered} {\mathbf{F}}_{\mathbf{B}}\mathbf{=}\mathbf{q}\left(v\times B\right) \\ \mathbf{F}\mathbf{=}\mathbf{qvBsin\theta } \end{gathered}$$

Where q is the charge of the particle, V is the velocity and B is the magnetic field

We know that the magnetic field force is given by

$$\begin{gathered} \mathrm{F}=\mathrm{q}\left(\mathrm{v}\times \mathrm{B}\right) \\ \mathrm{F}=\mathrm{q}\left|\mathrm{v}\right|\left|\mathrm{B}\right|\mathrm{sin\theta } \end{gathered}$$

By the second law of motion we know that the force on a body is

$\stackrel{\rightharpoonup }{\mathrm{F}}=\mathrm{m}\stackrel{\rightharpoonup }{\mathrm{a}}$

And the magnetic field force is given by

$\mathrm{F}=\mathrm{q}\left(\mathrm{v}\times \mathrm{B}\right)$

Equating the two we get

$$\begin{gathered} m\stackrel{\rightharpoonup }{\mathrm{a}}=\mathrm{q}\left(\mathrm{v}\times \mathbf{B}\right) \\ \stackrel{\rightharpoonup }{\mathrm{a}}=\frac{\mathrm{q}\left(\mathrm{v}\times \mathbf{B}\right)}{\mathrm{m}} \\ \stackrel{\rightharpoonup }{\mathrm{a}}=\frac{\mathrm{q}\left|\mathrm{v}\right|\left|\mathrm{B}\right|\mathrm{sin\theta }}{\mathrm{m}} \end{gathered}$$

For minimum acceleration we get when,$\mathrm{\theta }=0°\mathrm{we}\mathrm{have}{\mathrm{a}}_{\min }=0.$

For maximum acceleration we get,$\theta =90°$

Substitute all the value in the above equation

$$\begin{gathered} \overrightarrow{\mathrm{a}}=\frac{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(1.40\times {10}^6\mathrm{m}/\mathrm{s}\right)\left(7.40\times {10}^{-2}\mathrm{T}\right)}{9.11\times {10}^{-31}\mathrm{Kg}} \\ \overrightarrow{\mathrm{a}}=1.82\times {10}^{16}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the maximum acceleration is$a_{max}=1.82\times {10}^{16}\mathrm{m}/{\mathrm{s}}^2$

When the actual acceleration of the electron is one-fourth of the largest magnitude ${\mathrm{a}}^{\text{'}}=\frac{\mathrm{a}}{4}$, then the magnetic force on the electron is also one-fourth of the largest force.

${\mathrm{F}}^{\text{'}}=\frac{\mathrm{F}}{4}$

We get

$$\begin{gathered} \mathrm{qvBsin\varphi }=\frac{\mathrm{qvB}}{4} \\ \mathrm{sin\varphi }=\frac{1}{4} \\ \mathrm{\varphi }=14.5^{°} \end{gathered}$$

Hence the angle is$\mathrm{\varphi }=14.5°$