A positive charge q is projected in magnetic field of width mv 2 qB...

Referring the diagram,&

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

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

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

Now in triangle OAB,

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

$= \dfrac{1}{2}$

$⇒ sin\theta = \sin 45 °$

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

Time period for angle $2\pi$ is $T = \dfrac{2\pim}{qB}$

$∴$ Time by charged particle for $\theta = \dfrac{\pi}{4}$ in order to emerge from magnetic field is&

$T = \dfrac{\pim}{4qB}$.