Question

An electron of kinetic energy of $7.2\times {10}^{-18}J$ is revolving on circular path in magnetic field $9\times {10}^{-5}Wb/m^2$ then radius of its circular path is :-

• A. 1.25 cm
• B. 2.5 m
• C. 25 cm
• D. 25.0 cm

Answer 1

The correct option is C 25 cm

Given:

$k=7.2\times {10}^{-18}kg{m}^2/s^{2}B=9\times {10}^{-5}T$

Mass of electron$=m=9.1\times {10}^{-31}kg$

Charge of electron$=e=1.6\times {10}^{-19}C$

$R=\frac{\sqrt{2km}}{eB}=\frac{\sqrt{2\times 7.2\times {10}^{-18}kg{m}^2/s^2\times 9.1\times {10}^{-31}kg}}{1.6\times {10}^{-19}C\times 9\times {10}^{-5}T}=25cm$