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 $ heta$ through which the proton deviates from the initial direction of its motion is

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 $	heta$ through which the proton deviates from the initial direction of its motion is

Dr. Abubacker · Accepted Answer

(b) According to the following figure, $\sin	heta = \frac{d}{r}$
Also $r = \frac{\sqrt{2mk}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$
$	herefore\sin	heta = Bd\sqrt{\frac{q}{2mV}}$
$= 0.51 	imes 0.1\sqrt{\frac{1.6 	imes 10^{- 19}}{2 	imes 1.67 	imes 10^{- 27} 	imes 500 	imes 10^{3}}}$
$= \frac{1}{2} \Rightarrow 	heta = 30$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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