Uniform magnetic field B exists in a circular region of radius R. A particle having specific charge $\frac{q}{m}$ enters the region with velocity $v_{0}$ as shown. For $v_{o} = \frac{RqB}{α m s i n θ}$ the particle passes through the center O of the region. Fill the value of $α$ .

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The particle moves on a circle of radius $r = \frac{m v _{0}}{qB}$

It enters the region at P and must pass through the centre O, with PO = R. PO is a chord of the circular trajectory.

The velocity at P makes angle θ with PO, so the chord length is $PO = 2 r sin θ$

$R = 2 r sin θ \Rightarrow r = \frac{R}{2 s i n θ}$

$v_{0} = \frac{qB r}{m} = \frac{RqB}{2 m s i n θ}$

Comparing with $v_{0} = \frac{RqB}{α m s i n θ}$ : α = 2. Answer: 2

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