A 10g bullet having a charge of $4.00 μ C$ is fired at a speed of $270 m/s$ in a horizontal direction. A vertical magnetic field of $500 μ T$ exists in the space. The deflection of the bullet due to the magnetic field as it travels through 100 m is (Make appropriated approximations)

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$6.2 \times 10^{- 6} m$

None

$3.7 \times 10^{6} m$

$3.7 \times 10^{- 6} m$

answer is B.

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Detailed Solution

$R = \frac{mv}{Bq} \to (1) 2 θ = \frac{l}{R} = \frac{x}{R} since' θ' is small x = 100 m ∴ 2 θ = \frac{100}{R} θ = \frac{50}{R} \to (2) \tan θ = \frac{y}{100} \tan θ = θ θ = \frac{y}{100} \to (3) From (2) & (3) \frac{50}{R} = \frac{y}{100} \Rightarrow y = \frac{5000}{R} = \frac{5000 \times Bq}{mv} = \frac{5000 \times 500 \times 10^{- 6} \times 4 \times 10^{- 6}}{10 \times 10^{- 3} \times 270} = \frac{10^{7} \times 10^{- 6} \times 10^{- 6}}{10^{- 3} \times 27} = \frac{1}{27} \times 100 \times 10^{- 6} = 3.7 \times 10^{- 6} m$