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An electron beam travels with a velocity 1.6 × 10⁷ ms^–1, perpendicularly to magnetic field of intensity 0.1T. The radius of the path of the electron beam (m_e = 9 × 10^–31 kg)

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9 × 10^-5 m

9 × 10^-2 m

9 × 10^-4 m

9 × 10^-3 m

answer is C.

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Detailed Solution

$r = \frac{mv}{Bq} = \frac{9 \times 10^{- 31} \times 1.6 \times 10^{7}}{(0.1) (1.6 \times 10^{- 19})} = 9 \times 10^{- 4} m$

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