A particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_1$ at an angle $\alpha$ and leaves the field at $C$ with speed $v_2$ at an angle $\beta$ as shown. Then
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(d)
$\beta=\alpha$
$v_1=v_2\left(\because F_m \perp v\right)$
$T =\frac{2(\pi-\alpha) m}{q B}$
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