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 $through which the proton deviates from the initial direction of its motion is......$^o$
✓
Answer
(b)According to following figure $\sin \theta = \frac{d}{r}$
also $r = \frac{{\sqrt {2mk} }}{{qB}} = \frac{1}{B}\sqrt {\frac{{2mV}}{q}} $
$\therefore $ $\sin \theta = Bd\sqrt {\frac{q}{{2mV}}} $
$ = 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^o}$
also $r = \frac{{\sqrt {2mk} }}{{qB}} = \frac{1}{B}\sqrt {\frac{{2mV}}{q}} $
$\therefore $ $\sin \theta = Bd\sqrt {\frac{q}{{2mV}}} $
$ = 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^o}$
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