Question

A particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\overrightarrow{B}$ $($perpendicular to paper inward$)$ at $P$ with a velocity $v_0$ at an angle $\alpha$ and leaves the field at $Q$ with velocity $v$ at angle $\beta$ as shown in the figure :

• A. $\alpha =\beta$
• B. $v=v_0$
• C. $PQ=\frac{2m{v}_0\sin \alpha }{Bq}$
• D. The particle remains in field for time $t=\frac{2m(\pi -\alpha )}{Bq}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\alpha =\beta$
B $v=v_0$
C The particle remains in field for time $t=\frac{2m(\pi -\alpha )}{Bq}$
D $PQ=\frac{2m{v}_0\sin \alpha }{Bq}$

$a)$ As the magnetic field is uniform, the charge comes out symmetrically. Hence, $\alpha =\beta$

$b)$ $v=v_0$ since the speed does not change during motion in a plane perpendicular to the magnetic field

$c)r=\frac{m{v}_0}{qB}$

Length of arc $PQ:2r\sin \alpha =\frac{2m{v}_0\sin \alpha }{qB}$

d) $T=\frac{2\pi m}{qB}$

Time spent inside magnetic field is

$T\times \frac{(2\pi -2\alpha )}{2\pi }=\frac{2\pi m}{qB}\times \frac{(2\pi -2\alpha )}{2\pi }=\frac{2m}{qB}(\pi -\alpha )$