A particle of mass 10 mg and having a charge of 50mC is projected with a speed of 15 m/s into a uniform magnetic field of 125mT. Assuming that the particle is projected with its velocity perpendicular to the magnetic field, the time after which the particle reaches its original position for the first time is

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14 s

12 s

10 s

8 s

answer is C.

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Detailed Solution

Given,

Mass of particle, $m = 10 m g = 10 \times 10^{- 6} k g$

Charge of the particle, $q = 50 \times 10^{- 6} C$

Speed of the particle, $v = 15 m / s$

Magnetic Field, $B = 125 \times 10^{- 3} T$

Time period, $T = \frac{2 πm}{q B}$

$= \frac{2 (3.14) (10 \times 10^{- 6})}{(50 \times 10^{- 6}) (125 \times 10^{- 3})} = \frac{6.28 \times 10}{625 \times 10^{- 2}} = 10.05 s e c$

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