Physics - Magnetism Question with Solution | TestHub

Referring the diagram, 

width of magnetic field $$$w=\frac{mv}{\sqrt{2}qB}$

Also we know that when a charge is projected in perpendicular magnetic field then radius of path is 

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

Now in triangle OAB,

$\sin \theta =\frac{w}{R}=\frac{\frac{mv}{\sqrt{2}qB}}{\frac{mv}{qB}}$

$=\frac{1}{\sqrt{2}}$

$\Rightarrow \sin \theta =\sin 45°$

$=\sin \frac{\pi }{4}$

Time period for angle $2\pi$ is $T=\frac{2\pi m}{qB}$

$\therefore$ Time by charged particle for $\theta =\frac{\mathrm{\pi }}{4}$in order to emerge from magnetic field is 

$T=\frac{\pi m}{4qB}$.