Question

A charged particle having charge $q$, accelerated through a potential difference of <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $V\text{volt}$ enters a transverse magnetic field region in which it experiences a force $F$. If $V$ is increased to $5V$, the particle will experience a force.

• A. $F$
• B. $5F$
• C. $\frac{F}{5}$
• D. $\sqrt{5}F$

Answer 1

The correct option is D $\sqrt{5}F$

When a positive charged particle is accelerated by $V\text{volt}$, the loss in its electrostatic P.E. converts to gain in its K.E.

$qV=\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{\frac{2qV}{m}}$

$\because$ It enters into transverse magnetic field region (i.e.,) $\overrightarrow{v}$ to $\overrightarrow{B}$,


Thus, magnetic force experienced by the particle is,

$\overrightarrow{F_m}=q(\overrightarrow{v}\times \overrightarrow{B})=qvB(\because \theta ={90}^{\circ })$

Thus, $F=qvB$

$\Rightarrow F\propto v(\because v\propto \sqrt{V})$

$\therefore F\propto \sqrt{V}$

If $V$ changed to $5V$

$\frac{F_2}{F_1}=\sqrt{\frac{5V}{V}}=\sqrt{5}$

$\therefore F_2=\sqrt{5}{F}_1=\sqrt{5}F$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(d\right)$ is correct.