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
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
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Magnetic field strength, B =6.5 x 10-4 T
Charge of the electron, e = 1.6 x 10-19 C
Mass of the electron, me = 9.1 x 10-31 kg
Velocity of the electron, v = 4.8 x 106 m/s
Radius 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.