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A particle of charge q and mass m is projected with a velocity v₀ toward a circular region having uniform magnetic field B perpendicular and into the plane of paper from point P as shown in figure. R is the radius and O is the center of the circular region. If the line OP makes an angle $θ$ with the direction of v₀ then the value of v₀ so that particle passes through O is

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$\frac{qBR}{msin θ}$

$\frac{qBR}{2 msin θ}$

$\frac{2 qBR}{msin θ}$

$\frac{3 qBR}{2 msin θ}$

answer is B.

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Detailed Solution

If the particle passes through O, the situation can be shown in the figure below. Let r the radius of circular path then from the given figure

$\begin{array}{l} \sin θ = \frac{R / 2}{r} = \frac{R}{2 r}, r = \frac{mv}{qB} \\ \sin θ = \frac{RqB}{2 mv}; v_{0} = \frac{qBR}{2 msin θ} \end{array}$

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