Question

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$\mathbf{A}\mathbf{}\mathbf{to}\mathbf{}\mathbf{B}\mathbf{.}$

Short answer

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

(b) The time required for the electron to move from$\mathrm{A}\mathrm{to}\mathrm{B}\mathrm{is}1.114\times {10}^{-7}\mathrm{s}$

Step-by-step solution

Definition of magnetic field

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

Determine the magnetic field

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$\mathrm{A}\mathrm{to}\mathrm{B}$.

Determine the time  

The required for this displacement is calculated as

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