A particle moving with velocity v having specific charge (q/m) enters a region of magnetic field B having width $d = \frac{3 mv}{5 qB}$ . at angle 53° to the boundary of magnetic field. Find the angle $θ$ in the diagram.

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none

37°

90°

60°

answer is C.

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Detailed Solution

r = Radius of circular
path = mv/qB
$d = \frac{3 r}{5}$

From the diagram,

$\begin{array}{l} OM - ON = d \\ rsin 37^{\circ} - rsin (90^{\circ} - θ) = d \\ \Rightarrow \frac{3}{5} - \cos θ = \frac{3}{5} \\ \Rightarrow θ = 90^{\circ} \end{array}$

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