Question

A particle of charge $q$ and mass $m$ enters normally $($at point $P)$ in a region of magnetic field with speed $V$. It comes out normally from $Q$ after time $T$ as shown in the figure. The magnetic field $B$ is present only in the region of radius $R$ and is uniform. Initial and final velocities are along radial direction, and they are perpendicular to each other. For this to happen, which of the following expression(s) is/are correct?

• A. $B=\frac{mV}{qR}$
• B. $T=\frac{\pi R}{2V}$
• C. $T=\frac{\pi m}{2qB}$
• D. $B=\frac{mV}{2qR}$

Answer 1

Answer: (A), (B), (C)

$T=\frac{\pi m}{2qB}$

The path of the particle will have radius equal to $r$ and its centre at point $C$.


Centripetal force is provided by magnetic force.

So,

$\frac{m{V}^2}{r}=qVB$

$\Rightarrow r=\frac{mV}{qB}$

From figure, quadrilateral $POQC$ is a square.

$\therefore r=R$

Hence,

$R=\frac{mV}{qB}$

$\Rightarrow B=\frac{mV}{qR}...\left(i\right)$

$[$Option $\left(A\right)$ is correct$]$

Now,

$T=\frac{2\pi r/4}{V}=\frac{\pi R}{2V}$

$[$Option $\left(B\right)$ is correct$]$

Also,

$T=\frac{\pi R\times m}{2\times BqR}\left[\text{From}\right(i\left)\right]$

$\Rightarrow T=\frac{\pi m}{2qB}$

$[$Option $\left(C\right)$ is correct$]$