Question

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

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

Answer 1

The correct option is C ${30}^{\circ }$

Here, $\theta$ is the angle of emergence, and we know that

$\sin \theta =\frac{d}{R}$

Where, $R=\frac{\sqrt{2mV}}{\sqrt{q}B}$

Given that
$V=500\times {10}^3\text{V};B=0.51\text{T}$
And for proton,
$m=1.6\times {10}^{-27}\text{kg};q=1.6\times {10}^{-19}\text{C}$

$R=\frac{\sqrt{2\times 1.6\times {10}^{-27}\times 500\times {10}^3}}{\sqrt{1.6\times {10}^{-19}}\times 0.51}$

$R=0.2\text{m}$

$\therefore \sin \theta =\frac{10\times {10}^{-2}}{0.2}=\frac{1}{2}$

$\Rightarrow \theta ={\sin }^{-1}(\frac{1}{2})={30}^{\circ }$

Hence, option $\left(c\right)$ is correct.