Physics - Magnetism Question with Solution | TestHub

The radius of the circular path of a charged particle moving perpendicular to a uniform magnetic field is given by

$R=\frac{mv}{qB}$

When the speed is halved, the new speed is $v^{\prime }=\frac{v}{2}$

When the magnetic field is doubled, the new magnetic field is $B^{\prime }=2B$

The new radius $R^{\prime }$ is

$R^{\prime }=\frac{m{v}^{\prime }}{q{B}^{\prime }}=\frac{m\left(\frac{v}{2}\right)}{q(2B)}$

$R^{\prime }=\frac{1}{4}\left(\frac{mv}{qB}\right)=\frac{R}{4}$

Answer: $\frac{\mathrm{R}}{4}$