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
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10⁻⁴ T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s⁻¹ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10⁻¹⁹ C, me = 9.1×10⁻³¹ kg)
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Magnetic field strength, B = 6.5 G = 6.5 x 10⁻4 T
Speed of the electron, v = 4.8 x 106 m/s
Charge on the electron, e = 1.6 x 10-19 C
Mass of the electron, me = 9.1 x 10 –31 kg
Angle 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 106 )/( 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.