Question

A proton accelerated by a potential $V=500$ $KV$ moves through a transverse magnetic field $B=0.51$ $T$ as shown in the figure. Then, the angle $\theta$ through which the proton deviates from the initial direction of its motion is (approximately)

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

Answer 1

Answer: (B)

${30}^o$

Proton is accelerated by a potential difference $V=500kV$
So,
$\frac{1}{2}{mv}^2=qV$
$v=\sqrt{\frac{2qV}{m}}$
From the figure
$d=r\sin \theta$
$10\times {10}^{-2}=\sin \theta \times \frac{mv}{qB}$
${10}^{-1}=\frac{m\sin \theta }{qB}\sqrt{\frac{2qV}{m}}$

$\sin \theta =0.51$
$\theta ={30}^{\circ }$