JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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
- A \(qV\,\sqrt {\frac{{Bd}}{{2m}}} \)
- B \(\frac{B}{2}\sqrt {\frac{{qd}}{{mV}}} \)
- C \(\frac{B}{d}\sqrt {\frac{{q}}{{2mV}}} \)
- D \(Bd\sqrt {\frac{q}{{2mV}}} \)
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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