Question

A proton accelerated by a potential difference $500\, kV$ moves though a transverse magnetic field of $0.51 \,T$ as shown in figure. The angle $\theta$ through which the proton deviates from the initial direction of its motion is

• A. $15^{\circ}$
• B. $30^{\circ}$
• C. $45^{\circ}$
• D. $60^{\circ}$

Answer 1

Answer: (B)

According to following figure
$\sin \theta=\frac{d}{r}$
also $r=\frac{\sqrt{2 m k}}{q B}=\frac{1}{B} \sqrt{\frac{2 m V}{q}}$
$\therefore \sin \theta=B d \sqrt{\frac{q}{2 m V}}$
$=0.51 \times 0.1 \sqrt{\frac{1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times 500 \times 10^{3}}}=\frac{1}{2}$
$\Rightarrow \theta=30^{\circ}$