Question

# In Exercise 4.11 obtain the frequency of revolution of the electron inits circular orbit. Does the answer depend on the speed of the electron? Explain.

Solution

## Given: The magnetic field strength is 6.5× 10 −4  T, the charge of the electron 1.6× 10 −19  C, the velocity of the electron is 4.8× 10 6  m/s, the radius of the orbit is 4.2 cm, the mass of the electron is 9.1× 10 −31  kg. The velocity of the electron is given as, υ=rω Where, the angular frequency is ω and the radius of the orbit is r. The angular frequency of the electron is given as, ω=2πν Where, the frequency of revolution of the electron is ν. Since, in the circular orbit, centripetal force balances the magnetic force on the electron in the circular orbit. The balance of these forces is given as, m υ 2 r =eυB Further simplifying the above expression, we get eB= mυ r eB= m( ωr ) r ν= eB 2πm Where, the magnetic field strength is B, the charge of the electron is e and the mass of the electron is m. By substituting the given values in the above expression, we get ν= 6.5× 10 −4 ×1.6× 10 −19 2×3.14×9.1× 10 −31 =18.2× 10 6  Hz =18.2 MHz Thus, the frequency of the electron is 18.2 MHzand it is independent of the speed of the electron. PhysicsPhysics Part-I - NCERTStandard XII

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