Class 12 Physics

CBSE and UP Board

Moving Charges and Magnetism

Chapter-4 Exercise 4.11

NCERT Solutions Class 12 Physics Chapter 4 Question 11

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Magnetic field strength, B = 6.5 G = 6.5 x 10⁻

^{4}TSpeed of the electron, v = 4.8 x 10

^{6}m/sCharge on the electron, e = 1.6 x 10

^{-19}CMass of the electron, m

_{e}= 9.1 x 10^{–}^{31}kgAngle between the shot electron and magnetic field, 0 = 90°

Magnetic force exerted on the electron in the magnetic field is given as:

F = evB sin0

This force provides centripetal force to the moving electron. Hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron,

Fe = mv/r²

In equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

Fe = F

=> mv/r² = evB sin0

=> r = mv/ evB sin0

So,

r = (9.1 x 10

^{–}^{31 }x 4.8 x 10^{6})/( 6.5 x 10⁻^{4}x 1.6 x 10^{-19}x Sin 90º)= 4.2 x 10⁻² m = 4.2 cm

Hence ,the radius of the circular orbit of the electron is 4.2 cm.