An electron is moving in a cyclotron at a speed of $ 3.2 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1} $ in a magnetic field $ 5 \times 10^{-4} \mathrm{~T} $ perpendicular to it. The frequency of this electron is: $ \left(q=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right) $

Question

An electron is moving in a cyclotron at a speed of $ 3.2 \times 10^{7} \mathrm{~m} \mathrm{~s}^{-1} $ in a magnetic field $ 5 \times 10^{-4} \mathrm{~T} $ perpendicular to it. The frequency of this electron is: $ \left(q=1.6 \times 10^{-19} \mathrm{C}, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right) $, $ 1.4 \times 10^{5} \mathrm{~Hz} $, $ 1.4 \times 10^{7} \mathrm{~Hz} $, $ 1.4 \times 10^{6} \mathrm{~Hz} $, $ 1.4 \times 10^{9} \mathrm{~Hz} $

MARKS App · Accepted Answer

v = 3.2 × 10 7 m s - 1 ; B = 5 × 10 - 4 T The frequency of electron is f = q B 2 π m = 1.6 × 10 - 19 × 5 × 10 - 4 2 × 3.14 × 9.1 × 10 - 31 f = 1.4 × 10 7 H z = 14 M H z

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