91Ó°ÊÓ

The magnetic field force is given by

$$\begin{gathered} {\mathbf{F}}_{\mathbf{B}}\mathbf{=}\mathbf{q}\left(v×B\right) \\ \mathbf{F}\mathbf{=}\mathbf{\pm ç\pm ¹\mu þ²õ¾\pm ²Ôθ} \end{gathered}$$

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

$$\begin{gathered} \mathrm{F}=\mathrm{q}\left(\mathrm{v}×\mathrm{B}\right) \\ \mathrm{F}=\mathrm{q}\left|\mathrm{v}\right|\left|\mathrm{B}\right|\mathrm{²õ¾\pm ²Ôθ} \end{gathered}$$

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

$\stackrel{â‡\P Ä}{\mathrm{F}}=\mathrm{m}\stackrel{â‡\P Ä}{\mathrm{a}}$

And the magnetic field force is given by

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

Equating the two we get

$$\begin{gathered} m\stackrel{â‡\P Ä}{\mathrm{a}}=\mathrm{q}\left(\mathrm{v}×\mathbf{B}\right) \\ \stackrel{â‡\P Ä}{\mathrm{a}}=\frac{\mathrm{q}\left(\mathrm{v}×\mathbf{B}\right)}{\mathrm{m}} \\ \stackrel{â‡\P Ä}{\mathrm{a}}=\frac{\mathrm{q}\left|\mathrm{v}\right|\left|\mathrm{B}\right|\mathrm{²õ¾\pm ²Ôθ}}{\mathrm{m}} \end{gathered}$$

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

For maximum acceleration we get,$\mathrm{θ}=90°$

Substitute all the value in the above equation

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

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

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

$$\begin{gathered} \mathrm{\pm ç\pm ¹\mu þ²õ¾\pm ²ÔÏ•}=\frac{\mathrm{qvB}}{4} \\ \mathrm{²õ¾\pm ²ÔÏ•}=\frac{1}{4} \\ \mathrm{Ï•}=14.5^{°} \end{gathered}$$

Hence the angle is$\mathrm{ϕ}=14.5°$