An electron accelerated through a potential difference $V$, passes through a uniform transverse magnetic…
An electron accelerated through a potential difference $V$, passes through a uniform transverse magnetic field and experiences a force $F$. If the accelerating potential is increased to $2 \mathrm{~V}$, the electron in the same magnetic field will experience a force.
$F$
$\frac{F}{2}$
$\sqrt{2} F$
$2 F$
Solution
$\because$ Magnetic force,
$
\begin{aligned}
F & =q(\mathbf{v} \times \mathbf{B}) \\
& =q v B \sin \theta
\end{aligned}
$
For uniform transverse magnetic field, $\theta=90^{\circ}$
So,
$
F=q v B=q\left(\sqrt{\frac{2 q V}{m_e}}\right) B
$
when electron accelerated through a potential difference,
When electron accelerated through a potential difference $2 V$.
$
F^{\prime}=e\left(\sqrt{\frac{2 e(2 V)}{m_e}}\right) B
$
From Eqs. (i), we get $F^{\prime}=\sqrt{2} F$
