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^o to the boundary of magnetic field. Find the angle $θ$ in the diagram.

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37^o

90^o

60^o

None

answer is C.

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Detailed Solution

The tangent is making angles 53 and with y-axis. So normal will make angles 37 and 90- $θ$ with the y-axis. $\frac{mv}{qB} (\sin 37 - \cos θ) = \frac{3 mv}{6 qB} . So \cos θ = 0 . Hence θ = 90$

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