Electron of mass ' $m$ ' and charge ' $q$ ' is travelling with speed ' $v$ ' along a circular path of radius…
Electron of mass ' $m$ ' and charge ' $q$ ' is travelling with speed ' $v$ ' along a circular path of radius ' $R$ ' at right angles to a uniform magnetic field of intensity ' $B$ '. If the speed of the electron is halved and the magnetic field is doubled, the resulting path would have radius
$\frac{\mathrm{R}}{2}$
$\frac{\mathrm{R}}{4}$
2 R
4 R
Solution
From cyclotron motion and uniform circular motion,
i.e., $\mathrm{q} v \mathrm{~B} \sin 90^{\circ}=\frac{\mathrm{mv}^2}{\mathrm{R}}$
$\therefore \quad \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$...(i)
Given $\mathrm{v}^{\prime}=\frac{\mathrm{v}}{2}$ and $\mathrm{B}^{\prime}=2 \mathrm{~B}$
$\begin{aligned}
& \therefore \quad R^{\prime}=\frac{\mathrm{mv}^{\prime}}{\mathrm{qB}^{\prime}}=\frac{\mathrm{m} \frac{\mathrm{v}}{2}}{\mathrm{q} 2 \mathrm{~B}}=\frac{1}{4} \frac{\mathrm{mv}}{\mathrm{qB}} \\
& \quad \Rightarrow R^{\prime}=\frac{1}{4} R
\end{aligned}$
...[From(i)]