Question

An electron of mass $m=9.1\times {10}^{-31}\text{kg}$ charge $q=-1.6\times {10}^{-19}\text{C}$ experiences no deflection , if subjected to an electric field of $E=3.2\times {10}^5\text{V}/\text{m}$ and a magnetic field of $B=2.0\times {10}^{-3}\text{Wb}/{\text{m}}^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius ?

• A. $45\text{m}$
• B. $4.5\text{m}$
• C. $0.45\text{m}$
• D. $0.045\text{m}$

Answer 1

The correct option is C $0.45\text{m}$

Initial condition:


Where,
$\overrightarrow{v}\to \text{velocity}\overrightarrow{E}\to \text{electric field}\overrightarrow{B}\to \text{magnetic field}$

According to question, no deflection of charge occurs,

$\therefore$ Either ${\overrightarrow{F}}_{net}=0$ or ${\overrightarrow{F}}_{net}$ is along $\overrightarrow{v}$

Since according to the arrangement mentioned in question (see diagram)

${\overrightarrow{F}}_{net}$ cannot be along $\overrightarrow{v}$

$\therefore {\overrightarrow{F}}_{net}=0$

Using Lorentz's force equation,

$F=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

$\Rightarrow 0=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

$\Rightarrow \overrightarrow{E}=-\overrightarrow{v}\times \overrightarrow{B}$

$\therefore |v|=\frac{E}{B}=\frac{3.2\times {10}^5}{2\times {10}^{-3}}=1.6\times {10}^8\text{m}/\text{s}$

When $\overrightarrow{E}$ is removed $\therefore {\overrightarrow{F}}_{net}=q(\overrightarrow{v}\times \overrightarrow{B})$

$\Rightarrow |{\overrightarrow{F}}_{net}|=qvB$ (responsible for circular motion)

$qvB=\frac{m{v}^2}{R}\Rightarrow R=\frac{mv}{qB}$

$\Rightarrow R=\frac{9.1\times {10}^{-31}\times 1.6\times {10}^8}{1.6\times {10}^{-19}\times 2\times {10}^{-3}}$

$\therefore R=0.455\text{m}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.

Why this question ? To familiarize oneself how to use Lorentz force equation in different scenario's and to use radius formula for charge moving in magnetic field.