Question

An electron moving at right angle to a uniform magnetic field completes a circular orbit in $1micro-second$. Find the magnetic field.
Ans. $3.53\times {10}^{-5}T$

Answer 1

We know that time of revolution of a charge moving perpendicular in magnetic field is given by, $T=\frac{2\pi m}{qB}$

here $m=9.1\times {10}^{-31}kg,q=e=1.6\times {10}^{-19}C$

so, we get , magnetic field, $B=\frac{2\pi m}{qT}=\frac{2\times 3.14\times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}\times 10-6}$

solving it we get, $B=3.53\times {10}^{-5}T$