An electron at point in figure has a speed ${\mathbf{v}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{41}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$. Find (a) the magnetic field that will cause the electron to follow the semicircular path from to and (b) The time required for the electron to move from$\mathbf{A}\mathbf{}\mathbf{to}\mathbf{}\mathbf{B}\mathbf{.}$

The term magnetic field may be defined as the area around the magnet behave like a magnet.

For reach point electron follow trajectory path and force acting on it is

$$\begin{gathered} \stackrel{\rightharpoonup }{{\mathrm{F}}_{\mathrm{B}}}=\mathrm{q}\left(\overrightarrow{{\mathrm{v}}_0}\times \stackrel{\rightharpoonup }{\mathrm{B}}\right) \\ \stackrel{\rightharpoonup }{{\mathrm{F}}_{\mathrm{B}}}=\left(-\mathrm{e}\right)\left(\overrightarrow{{\mathrm{v}}_0}\times \stackrel{\rightharpoonup }{\mathrm{B}}\right) \\ \left(1\right) \end{gathered}$$

According to the right-hand thumb rule magnetic field must point into the page because from the graph initial force must point to the right. And according to Second law of motion the magnitude of the force is product of mass and acceleration

${\mathrm{F}}_{\mathrm{B}}=\mathrm{ma}$

And radial acceleration is so $\frac{{\mathrm{v}}^2}{\mathrm{R}}\mathrm{so}$

${\mathrm{F}}_{\mathrm{B}}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$ … (2)

Now form (1) equation ${\overrightarrow{\mathrm{F}}}_{\mathrm{B}}=\mathrm{evBsin}\left(\mathrm{\theta }\right)$where $\mathrm{\theta }=90°\mathrm{So}{\mathrm{F}}_{\mathrm{B}}=\mathrm{evB}$

From (1) and (2)

$\mathrm{eVB}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}$

Put all the given values and calculate magnetic field

$$\begin{gathered} \mathrm{B}=\frac{\mathrm{mV}}{\left|\mathrm{q}\right|\mathrm{R}} \\ \mathrm{B}=\frac{\left(9.109\times {10}^{-31}\mathrm{kg}\right)\left(1.41\times {10}^6\mathrm{m}/\mathrm{s}\right)}{\left(1.602\times {10}^{-19}\mathrm{C}\right)\left(0.050\mathrm{m}\right)} \\ \mathrm{B}=1.60\times {10}^{-4}\mathrm{T} \end{gathered}$$

Hence, the magnetic field is $1.60\times {10}^{-4}\mathrm{T}$that will cause the electron to follow the semicircular path from$\mathrm{A}\mathrm{to}\mathrm{B}$.

The required for this displacement is calculated as

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{\pi R}}{{\mathrm{v}}_0} \\ \mathrm{t}=\frac{\mathrm{\pi }\times 0.050\mathrm{m}}{1.41\times {10}^6\mathrm{m}/\mathrm{s}} \\ \mathrm{t}=1.114\mathrm{s} \\ \mathrm{Hence},\mathrm{the}\mathrm{time}\mathrm{required}\mathrm{for}\mathrm{the}\mathrm{electron}\mathrm{to}\mathrm{move}\mathrm{from}\mathrm{A}\mathrm{to}\mathrm{Bis}1.114\times {10}^{-7}\mathrm{s} \end{gathered}$$