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A particle of charge q and mass m is projected with a velocity $v_{0}$ towards a circular region having uniform magnetic field B perpendicular and into the plane of paper from point P as shown in the figure. R is the radius and O is the centre of the circular region. If the line OP makes an angle $θ$ with the direction of $v_{0}$ , then the value of $v_{0}$ so that particle passes through O is

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$\frac{3 q B R}{2 m \sin θ}$

$\frac{q B R}{2 m \sin θ}$

$\frac{2 q B R}{m \sin θ}$

$\frac{q B R}{m \sin θ}$

answer is B.

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Detailed Solution

Particle moves on a circular path. Centre of circle lies on a vertical line passing through P. If radius of the circle is $r$ , then

$r = \frac{m v}{q B}$

$\sin θ = \frac{R}{2 r}$

So, $\frac{R}{2 \sin θ} = \frac{m v}{q B}$

$\Rightarrow v = \frac{q B R}{2 m \sin θ}$

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