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