Question

A charged particle of mass $ m $ and charge $ q $ moving perpendicular to a uniform magnetic field of strength $ B $ has kinetic energy $ E $ . The radius of curvature of its path is

• A. $ \sqrt{\frac{2Em}{B^{2}q^{2}}} $
• B. $ \sqrt{\frac{Em}{2B^{2}\,q^{2}}} $
• C. $ \frac{1}{2}\sqrt{\frac{Em}{B^{2}\,q^{2}}} $
• D. $ 2\sqrt{\frac{Em}{B^{2}\,q^{2}}} $

Answer 1

Answer: (A)

If $v$ is the velocity of the particle, then
$E=\frac{1}{2}mv^{2}$
or $V=\sqrt{\frac{2E}{m}} \ldots\left(i\right)$
As the particle describes a circular path of radius $ r$ in a perpendicular magnetic field,
$\therefore \frac{mv^{2}}{r}=qvB$
or $ r=\frac{mv}{qB}=\frac{m}{qB}\sqrt{\frac{2E}{m}}$
$=\sqrt{\frac{2Em}{q^{2}B^{2}}}$ (using (i))