A beam of electrons at rest is accelerated by a potential $V$. This beam experiences a force $F$ in a…
A beam of electrons at rest is accelerated by a potential $V$. This beam experiences a force $F$ in a uniform magnetic field. The accelerating field is increased to $V$ ' and the force experienced by the electrons in the same field is ' $2 F$ ' . The ratio is $\frac{V}{V^{\prime}}$
$2: 1$
$1: 2$
$1: 4$
$1: 1$
Solution
$\begin{aligned} & \text { K.E. } \frac{1}{2} m v^2=e V \\ & \therefore v=\sqrt{\frac{2 e V}{m}} \\ & F=e V B=e \sqrt{\frac{2 e V}{m}} B \\ & \text { Given } F^{\prime}=2 F=2 e \sqrt{\frac{2 e V}{m}} B \\ & \Rightarrow \frac{1}{2} e \sqrt{\frac{2 e V^{\prime}}{m}} B=e \sqrt{\frac{2 e V}{m}} B \\ & \Rightarrow V^{\prime}=4 V\end{aligned}$
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