Question

A particle of mass $M$ and positive charge $Q$ moving with a constant velocity $u_1=4\hat{i}\text{m/s}$, enters a region of uniform magnetic field normal to $xy-$plane. The region of the magnetic field extends from $x=0$ to $x=L$ and for all values of $y$. After passing through this the particle emerges on the other side after $10\text{ms}$ with a velocity $u_2=2(\sqrt{3}\hat{i}+\hat{j})\text{m/s}$. For the scenario described the direction and magnitude of magnetic field respectively are;

• A. $-vez-$direction, $\frac{30\pi M}{7Q}$
• B. $-vez-$direction, $\frac{50\pi M}{3Q}$
• C. $+vez-$direction, $\frac{100\pi M}{3Q}$
• D. $-vez-$direction, $\frac{50\pi M}{3Q}$

Answer 1

The correct option is B $-vez-$direction, $\frac{50\pi M}{3Q}$


Given:
The initial velocity is, $u_1=4\hat{i}\text{m/s}$
The velocity at emergence, $u_2=2\sqrt{3}\hat{i}+2\hat{j}$

The magnetic force acting on a moving charge in a magnetic field is given by

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

The direction of force $F$ acting on the particle will be along $+vey$-direction at the entry for shown deflection.

So at entrance, $\overrightarrow{V}=V\hat{i};\overrightarrow{F}=F\hat{j}$

So, from above expression, we get

$\overrightarrow{B}=B(-\hat{k})$

Thus, we get the direction of $\overrightarrow{B}$ as perpendicularly inwards $(\otimes )$ or along $-vez$-direction.

At the emergence,

$\tan \theta =\frac{v_y}{v_x}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\therefore \theta ={30}^{\circ }$

Thus, $\overrightarrow{u_2}$ makes angle $\frac{\pi }{6}$ from $+vex-$axis.

Angle rotated by the particle $=\frac{\pi }{6}$

$\therefore \omega t=\frac{\pi }{6}$

$(\frac{QB}{M})t=\frac{\pi }{6}$

$(\because t=10\times {10}^{-3}=\frac{1}{100}\text{s})$

$\therefore B=\frac{100\pi M}{6Q}=\frac{50\pi M}{3Q}\otimes$

Hence, option $\left(b\right)$ is the correct answer.