Question

A particle having a charge of $2.0\times {10}^{-8}$ C and a mass of $2.0\times {10}^{-10}$ g is projected with a speed of $2.0\times {10}^3$ m $s^{-1}$ in a region having a uniform magnetic field of 0.10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.

Answer 1

$q=2.0\times {10}^{-6}C$,

$B=0.10T$

$m=2.0\times {10}^{-10}$ kg

$=2\times {10}^{-13}$ kg,

$V=2.0\times {10}^3$ m/s

Using qVB = $\frac{m{V}^2}{r}$

So, radius $r=\frac{mV}{qB}=20$ cm

$\therefore T=\frac{2\pi r}{v}$

So, $T=6.28\times {10}^{-4}s$