Question

Electrons accelerated by a potential difference $V$ enter a uniform magnetic field of flux density $B$ at right angles to the field. They describe a circular path of radius $r$. If now $V$ is doubled and $B$ is also doubled, the radius of the new circular path is :

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

Answer 1

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

Radius of charged particle in magnetic field is,
$r=\frac{mv}{qB}$
$\frac{m{v}^2}{2}=qV$
$mv=\sqrt{2qVm}$
$r=\frac{mv}{qB}=\frac{\sqrt{2qVm}}{qB}=\frac{1}{B}\times (\sqrt{\frac{2Vm}{q}})$
$r^{\prime }=\frac{1}{2B}\times (\sqrt{\frac{2\times 2Vm}{q}})=\frac{r}{\sqrt{2}}$