Question

A proton accelerated by a potential difference $V,$ moves through a transverse magnetic field $B$ as shown in the figure. Then the angle $\theta $ through which the proton deviates from the initial direction of its motion is

• A. $\theta=\sin ^{-1}\left(d B \sqrt{\frac{q V}{2 m}}\right)$
• B. $\theta=\sin ^{-1}\left(d B \sqrt{\frac{q}{2 m V}}\right)$
• C. $\theta=\tan ^{-1}\left(d B \sqrt{\frac{q V}{2 m}}\right)$
• D. $\theta=\tan ^{-1}\left(d B \sqrt{\frac{q}{2 m V}}\right)$

Answer 1

Answer: (B)

$B q v=\frac{m v^{2}}{r}$
$\frac{1}{2} m v^{2}=V q$
And $\sin \theta=\frac{d}{r}$
$r=\frac{m v}{q B}=\frac{m}{q B} \sqrt{\frac{2 V q}{m}}$
$\sin \theta=d B \sqrt{\frac{q}{2 m V}}$
$\theta=\sin ^{-1}\left(d B \sqrt{\frac{q}{2 m V}}\right)$