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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)
Options:
Solution:
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Verified Answer
The correct answers are:
The particle enters Region III only if its velocity $v>\frac{q l B}{m}$
,
Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$
,
Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I
The particle enters Region III only if its velocity $v>\frac{q l B}{m}$
,
Path length of the particle in Region II is maximum when velocity $v=\frac{q I B}{m}$
,
Time spent in Region II is same for any velocity $v$ as long as the particle returns to Region I
$\overrightarrow{\mathbf{V}} \perp \overrightarrow{\mathbf{B}}$ 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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