A particle of mass ‘m ’ and charge ‘ q’ is projected into a region having a perpendicular uniform magnetic field ‘B’ of width d . Find the angle of deviation ‘ $θ$ ’ of the particle as it comes out of the magnetic field.

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$\sin^{- 1} (\frac{dqB}{mv})$

$\sin^{- 1} (\frac{Bq}{mvd})$

$\sin^{- 1} (\frac{mv}{dqB})$

$\sin^{- 1} (\frac{Bq}{mv})$

answer is A.

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Detailed Solution

The radius of the circular orbit is $r = \frac{m v}{q B} - - - (1)$
The deviation $' θ'$ may be obtained from $\sin θ = \frac{d}{r}$ $= \frac{d q B}{m v}$
$∴ θ = \sin^{- 1} (\frac{d q B}{m v})$ .

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