Q 12-04-020NEETJEE MainHard
A charged particle enters, at right angles, a region of uniform magnetic field of width $d$ (the field is perpendicular to the plane of motion). In the field its path would have radius $2d$. The angle through which the particle is deviated when it leaves the field region is
Answer: (A) $30^\circ$
Inside the field the particle moves on a circular arc of radius $r$. Crossing a strip of width $d$, the arc turns through angle $\theta$ with $\sin\theta = \dfrac{d}{r} = \dfrac{d}{2d} = \dfrac{1}{2}$.
So $\theta = 30^\circ$.
Solution by Sreeraj P, M.Sc Physics