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6 ELECTROMAGNETISM

P60 009

An electron moving with a velocity of 4.56×10⁶ m.s^–1 enters a uniform magnetic field of strength 3.21×10^-2 T at an angle of 42°.

(a) Describe the path of the electron traveling through the magnetic field.

(b) Calculate the radius of the path followed by the electron?

View solution below only after you have completed the answering the question.

Solution

right hand palm rule Þ

Magnetic force on charged particle is directed towards centre of a circle à circular motion

electron charge e = q = 1.602´10^-19 C

electron mass m_e = m = 9.11´10^-11 kg

electron velocity v = 4.56´10⁶ m.s^-1

magnetic field B = 3.21´10^-2 T

electron moving at right angles to B field q = 90^o sinq = 1

Force on charge moving in magnetic field

Magnetic force = Centripetal force F_c = m v² / R

R = m v / B q = 8.08´10^-4 m