An electron moves in a circular path of radius 20 cm in a uniform magnetic field of 2 x 10³ T Find the speed of the electron and period of revolution. Mass of electron = 9.1 x 10-31 kg. [Ans: 7.02 x 10 m/sec and 5.6 x 10 rev/sec]

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-08-16T19:42:37+02:00

The linear speed of the electron is 7.03 x 10¹³ m/s and the angular speed of the electron is 5.6 x 10¹³ rev/s.

The speed of the electron is calculated as follows;

F = ma = mv²/r ---(1)

F = qvB ----- (2)

mv²/r = qvB

mv/r = qB

mv = qBr

v = qBr/m

where;

v = (1.6 x 10⁻¹⁹ x 2 x 10³ x 0.2) / (9.1 x 10⁻³¹)

v = 7.03 x 10¹³ m/s

ω = v/r

ω = (7.03 x 10¹³ m/s) / (0.2 m) = 3.51 x 10¹⁴ rad/s

ω = 3.51 x 10¹⁴ rad/s x 1 rev/2π rad

ω = 5.6 x 10¹³ rev/s

Thus, the linear speed of the electron is 7.03 x 10¹³ m/s and the angular speed of the electron is 5.6 x 10¹³ rev/s.

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