An electron moves in a circular path of radius 0.01 m in a perpendicular magnetic field of \( 10^{-3}\,\mathrm{T} \). The speed of the electron is nearly (e/m of the electron \( =1.76\times10^{11}\,\mathrm{C\,kg^{-1}} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.76\times10^{6}\,\mathrm{m\,s^{-1}} \)
The field is perpendicular to the motion of the electron, so the magnetic force \( qvB \) provides the centripetal force. Then \( mv^{2}/r=qvB \), and the radius of the circular path is \( r=mv/qB \).
For the electron \( q=e \), so the speed is:
\( \begin{aligned} v &= \frac{e}{m}\,Br \\ &= 1.76\times10^{11}\,\mathrm{C\,kg^{-1}}\times10^{-3}\,\mathrm{T}\times0.01\,\mathrm{m} \\ &= 1.76\times10^{6}\,\mathrm{m\,s^{-1}} \end{aligned} \)