Question

An electron accelerated by a potential difference $V$ enters a uniform magnetic field of flux density $B$ at right angles to the field. It describes a circular path of radius $r$. If $V$ and $B$ are doubled, then the radius of the new circular path is

• A. $\frac{r}{\sqrt{2}}$
• B. $2\sqrt{2}r$
• C. $2r$
• D. $4r$

Answer 1

The correct option is A $\frac{r}{\sqrt{2}}$

As we know that,

Kinetic energy of electron, $KE=qV$

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

$v=\sqrt{\frac{2Vq}{m}}.......\left(1\right)$

Force on moving electron due to magnetic field is

$F_m=\frac{m{v}^2}{r}$

$Bqv\sin {90}^{\circ }=\frac{m{v}^2}{r}$

$r=\frac{mv}{Bq}=\frac{m}{Bq}\sqrt{\frac{2Vq}{m}}\left[\text{from eq.}\right(1\left)\right]$

$r=\sqrt{\frac{2Vq{m}^2}{m{B}^2{q}^2}}=\sqrt{\frac{2Vm}{B^{2}q}}$

For same charged particle,

$r\propto \frac{\sqrt{V}}{B}$

$\therefore \frac{r_2}{r_1}=\sqrt{\frac{V_2}{V_1}}\times \frac{B_1}{B_2}$

From question, $B_2=2B;V_2=2V$
$r_1=r;V_1=V;B_1=B$

$\frac{r_2}{r}=\sqrt{\frac{2V}{V}}\times \frac{B}{2B}=\frac{\sqrt{2}}{2}$

$\therefore r_2=\frac{r}{\sqrt{2}}$

Hence, option $\left(a\right)$ is correct.