A non-conducting cylindrical shell of radius R is placed in uniform magnetic field. The axis of the cylinder is parallel to the direction of magnetic field. A hole is drilled on the surface of the cylinder and a particle having charge q and mass m is projected with velocity $V_{0}$ perpendicular to magnetic field and directed towards the axis of cylinder shown in diagram. The particle collides elastically with inner wall of the cylinder and rebounds.

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The minimum number of collisions after which it will emerge is two.

The maximum speed $V_{0}$ for which it can come out is $V_{0} = (\frac{q B R}{m}) \sqrt{3}$

The minimum time in which it can come out will be $(\frac{π m}{q B})$

If the particle is projected with a velocity $\frac{V_{0}}{3} {w h e r e V_{0} = (\frac{q B R}{m}) \sqrt{3}}$ it will come out after five collisions.

answer is A, B, C, D.

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Detailed Solution

From fig. The minimum number of collisions are two collisions

$∴ θ = 60^{0}; t an θ = \frac{r}{R} = \sqrt{3}$

$\frac{m v_{0}}{B q R} = \sqrt{3} \Rightarrow V_{0} = [\frac{B q R}{m}] \sqrt{3}$

Question Image
For 360º turn in a magnetic field time is $T = \frac{2 π m}{B q}$
For 60º arc turn time is $[\frac{π m}{3 B q}]$
Required time for 3 truns $t = \frac{π m}{B q}$
$i f v = \frac{v_{0}}{3} = 1$

Question Image

$\tan θ = r / R = \frac{m v_{0}}{B q R} \Rightarrow \tan θ = \frac{m}{B q R} . \frac{v_{0}}{3} = \frac{m}{3 B q R} [\frac{B q R}{m}] \sqrt{3} = \frac{1}{\sqrt{3}} \Rightarrow θ = 30^{0}$

At centre are arc path of particle makes 60º at the centre of cylinder. The particle makes five collisions and came out from the field $(6 \times 60^{0} = 360^{0})$

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