JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2015
- Hard
- MCQ
A proton (mass \(m\) ) accelerated by a potential difference \(V\) flies through a uniform transverse magnetic field \(B.\) The field occupies a region of space by width \('d'\). If \(\alpha \) be the angle of deviation of proton from initial direction of motion (see figure), the value of \(sin\,\alpha \) will be
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(Bd\sqrt {\frac{q}{{2mV}}} \)
Step-by-step Solution
Detailed explanation
From figure, \(sin\,\alpha =dlR\)
And we know, \(\frac{m v^{2}}{R}=q v B\)
\(\Rightarrow \quad R=\frac{m v}{q B}\)
\(\because \sin \alpha=\frac{d q B}{m v}\)
\(\sin \alpha=B d \sqrt{\frac{q}{2 m V}}\left[\because q V=\frac{1}{2} m v^{2}\right]\)
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