Question

When two electrons enter into a magnetic field with different velocities, they deflect in different circular paths, in such a way that the radius of one path is double that of the other. $1\times {10}^7{ms}^{-1}$ is the velocity of the electron in smaller circle of radius $2\times {10}^{-3}m$. The velocity of electron in the other circular path is:

• A. $4\times {10}^7{ms}^{-1}$
• B. $4\times {10}^6{ms}^{-1}$
• C. $2\times {10}^7{ms}^{-1}$
• D. $2\times {10}^6{ms}^{-1}$

Answer 1

Answer: (C)

$2\times {10}^7{ms}^{-1}$

Balancing centripetal force on the electron with force due to magnetic field, we get,

$\frac{m{v}^2}{r}=qvB$

$r=\frac{mv}{qB}$
As $r^{\prime }=2r$; (Given)
Therefore, $v^{\prime }=2v$; where v' is velocity of electron in other circular path.

Since radius$\left(r\right)$ is proportional to the velocity $\left(v\right)$, hence, electron whose velocity is double will have radius double as that of the electron with lower velocity. Hence, velocity of second electron will be : $2\times 1\times {10}^{7}m/s=2\times {10}^{7}m/s$