Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.12

NCERT Solutions Class 12 Physics Chapter 4 Question 12

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Magnetic field strength, B =6.5 x 10

^{-4}TCharge of the electron, e = 1.6 x 10

^{-19}CMass of the electron, m

_{e}= 9.1 x 10^{-31}kgVelocity of the electron, v = 4.8 x 10

^{6}m/sRadius of the orbit, r = 4.2 cm = 0.042 m

Frequency of revolution of the electron = v

Angular frequency of the electron = ω = 2πv

Velocity of the electron is related to the angular frequency as: v = rω

In the circular orbit, the magnetic force on the electron is balanced by the centripetal force.

Hence, we can write:

mv²/r = evB=> eB = mv/r =m (rω)/r = m (r.2πv)/r

=> v = Be/2πm

This expression for frequency is independent of the speed of the electron.

On substituting the known values in this expression, we get the frequency as

v = ( 6.5 x 10

^{-4}x 1.6 x 10^{-19})/( 2 x 3.14 x 9.1 x 10^{-31}= 1.82 x 10⁶ Hz or approx. 18MHz

Hence ,the frequency of the electron is around 18MHz and is independent of the speed of the electron.