Question

A long straight wire carries a current of 50 A. An electron, traveling at1.0×107m/s, is 5.0 cm from the wire. What is the magnitude of the magnetic force on the electron if the electron velocity is directed (a) toward the wire, (b) parallel to the wire in the direction of the current, and (c) perpendicular to the two directions defined by (a) and (b)?

Answer 1

Answer:

Recall, The magnetic field on a electron in a wire, Bₓ = (µ₀ I)/(2πr)

Note: Magnetic force direction is at right angles to electron motion.

Magnetic force, F = q * v * Bₓ

Substituting Bₓ into F, we have

F = q * v * Bₓ = (q * v * µ₀* I)/(2π r) = (1.6 x 10⁻¹⁹ x 1.0 x 10⁷ x 4π x 10⁻⁷ x 50)/( 2π x 5 x 10⁻²) = 3.2 x 10⁻¹⁶ N

a) Hence, the magnitude of the magnetic force on the electron if the electron velocity is directed (a) toward the wire is 3.2 x 10⁻¹⁶ N  and opposite the current

b) Hence, the magnitude of the magnetic force on the electron if the electron velocity is directed (b) parallel to the wire in the direction of the current is 3.2 x 10⁻¹⁶ N  and away from the wire.

c) Hence, the magnitude of the magnetic forces on the electrons if the electron velocity is perpendicular to the two directions defined by (a) and (b) is 3.2 x 10⁻¹⁶ N  and (a) direction - opposite the current, (b) same direction as current.