A particle moving with velocity v having specific charge (q/m) enters a region of magnetic field B having width d=3mv5qB at angle 53° to the boundary of magnetic field. Find the angleθ in the diagram.

sinθ=dr=3mV5qB(mVBq) ⇒sinθ=35 ⇒θ=37°

sinθ=dr=3mV5qB(mVBq) ⇒sinθ=35 ⇒θ=37°

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

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37 °

60 °

90 °

none

answer is C.

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\sin θ = \frac{d}{r} = \frac{\frac{3mV}{5qB}}{(\frac{mV}{Bq})}

$\Rightarrow \sin θ = \frac{3}{5}$

$\Rightarrow θ = 37 °$

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