Question

A particle of mass $m$ and charge $q$ is projected into a region having a perpendicular magnetic field $B$. Find the angle of deviation as it comes out of the magnetic field if the width d of the region is very slightly less than $\frac{mv}{2qB}$

• A. ${30}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${45}^o$

Answer 1

The correct option is A ${30}^o$

$\frac{m{v}_{\perp }}{qB}=\frac{mv}{2qB}$
$\therefore \frac{mv\sin \theta }{qB}=\frac{mv}{2qB}$
$\Rightarrow \sin \theta =\frac{1}{2}=\sin {30}^0$
$\Rightarrow \theta ={30}^0$