Question

A particle of mass M and positive charge Q, moving with a constant velocity $u_1=4\hat{i}m{s}^{-1}$, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from $x=0$ to $x=L$ for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity ${\overrightarrow{u}}_2=2(\sqrt{3}\hat{i}+\hat{j})m{s}^{-1}$. The correct statement(s) is (are)

• A. the direction of the magnetic field is -z direction
• B. the direction of the magnetic field is +z direction
• C. the magnitude of the magnetic field $\frac{50\pi M}{3Q}$ units
• D. the magnitude of the magnetic field $\frac{100\pi M}{3Q}$ units

Answer 1

Answer: (A), (C)

The correct options are
A the direction of the magnetic field is -z direction
C the magnitude of the magnetic field $\frac{50\pi M}{3Q}$ units

Time period of circular motion $T=\frac{2\pi M}{QB}$

Time spent inside the magnetic field is $t=\frac{\pi }{6}\times T=\frac{\frac{\pi }{6}}{2\pi }\times \frac{2\pi M}{QB}=\frac{\pi M}{6QB}$

Give, $t=10ms=10\times {10}^{-3}s$

$\therefore \frac{\pi M}{6QB}=10\times {10}^{-3}$

$\therefore B=\frac{50\pi M}{QB}$
As seen from the figure, $\overrightarrow{u}=4\hat{i}$
Since the partcile bends upward, B has to be in $\hat{k}$ direction since $\overrightarrow{v}\times \overrightarrow{B}$ is in $\hat{j}$ direction.