JEE Advanced · Physics · 20. Magnetism & Current
A particle of mass \(m\) and charge \(q\), moving with velocity \(v\) enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field \(B\) perpendicular to the plane of the paper. The length of the Region II is \(l\). Choose the correct choice(s)
- A The particle enters Region III only if its velocity \(v>\frac{q l B}{m}\)
- B The particle enters Region III only if its velocity \(v < \frac{q I B}{m}\)
- C Path length of the particle in Region II is maximum when velocity \(v=\frac{q I B}{m}\)
- D Time spent in Region II is same for any velocity \(v\) as long as the particle returns to Region I
Answer & Solution
Correct Answer
(A) The particle enters Region III only if its velocity \(v>\frac{q l B}{m}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathbf{V}} \perp \overrightarrow{\mathbf{B}}\) in region II. Therefore, path of particle is circle in region II. Particle enters in region III if, radius of circular path, \(r>1\) or \(\frac{m v}{B q}>1\) or \(v>\frac{B q l}{m}\)
If \(v=\frac{B q l}{m}, r=\frac{m v}{B q}=l\), particle will turn back and path length will be maximum as shown in figure in region II. If particle returns to region I, time spent in region II will be \(t=\frac{T}{2}=\frac{\pi m}{B q}\), which is independent of \(v\).
\(\therefore\) correct options are (a), (c) and (d).
If \(v=\frac{B q l}{m}, r=\frac{m v}{B q}=l\), particle will turn back and path length will be maximum as shown in figure in region II. If particle returns to region I, time spent in region II will be \(t=\frac{T}{2}=\frac{\pi m}{B q}\), which is independent of \(v\).
\(\therefore\) correct options are (a), (c) and (d).
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