The region between $y = 0$ and $y = d$ contains a magnetic field $\vec{B} = B\hat{z}$. A particle of mass $m$ and charge $q$ enters the region with a velocity $\vec{v} = v\hat{i}$. If $d = \dfrac{mv}{2qB}$, the acceleration of the charged particle at the point of its emergence at the other side is:
Answer: (D) None of the above
The radius of the circular path is $r = \dfrac{mv}{qB} = 2d$.
At entry ($y = 0$, velocity $v\hat{i}$) the force on a positive charge is $q v\hat{i}\times B\hat{z} = -qvB\hat{j}$, i.e. away from the field region. The particle turns back, completes a semicircle and leaves through $y = 0$ with velocity $-v\hat{i}$, never reaching $y = d$.
At that point the acceleration is $\dfrac qm(-v\hat{i})\times B\hat{z} = \dfrac{qvB}{m}\hat{j}$, which is none of the given vectors.
(Even for a negative charge, which bends towards $+y$ and leaves at $y = d$ after turning through $60^\circ$, the acceleration is $\dfrac{|q|vB}{m}\left(-\dfrac{\sqrt3}{2}\hat{i} + \dfrac12\hat{j}\right)$, again not listed.)
Solution by Sreeraj P, M.Sc Physics