Question

An electron of $10$ eV energy is revolving around circular path in magnetic field of $1\times {10}^{-4}$ weber$/$ metre${}^2$.
Find (i) Speed of electron, (ii) radius of circular path, (iii) time period.

Answer 1

Magnetic field $B=1\times {10}^{-4}Wb/m^2$
Mass of electron $m=9.1\times {10}^{-31}kg$
Magnitude of charge of electron $e=1.6\times {10}^{-19}C$
Kinetic energy of electron $K=10eV$
(i) : Speed of electron $v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\times 10\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}}=5.93\times {10}^{6}m/s$
(ii) : Radius of circular path $R=\frac{mv}{eB}$
$⟹R=\frac{9.1\times {10}^{-31}\times 5.93\times {10}^6}{1.6\times {10}^{-19}\times {10}^{-4}}=33.73\times {10}^{-2}m=33.73cm$
(iii) : Time period $T=\frac{2\pi m}{eB}$
$⟹T=\frac{2\pi \times 9.1\times {10}^{-31}}{1.6\times {10}^{-19}\times {10}^{-4}}=35.72\times {10}^{-8}s$