An electron (charge = -1.6 x 10^-19C) is moving at 3 x 10^5 m/s in the positive x direction. A magnetic field of 0.8 T is in the positive z-direction. The magnetic force on the electron is:

Explanation

Magnetic Force(F)=q⋅(v×B) Where: q = charge of the particle = -1.6 x 10-19 C (Coulombs) v = velocity of the particle = 3 x 105 m/s in the positive x-direction B = magnetic field = 0.8 T in the positive z-direction F=(−1.6×10−19C)⋅(3×10 5m/s)×(0T,0T,0.8T) The cross product of v andB will give us a vector that represents the direction of the magnetic force. The magnitude of the magnetic force is given by ∣F∣=∣q∣⋅∣v∣⋅∣B∣⋅sin( θ ) ∣F∣=∣q∣⋅∣v∣⋅∣B∣⋅sin(θ), where θ is the angle between v and B.Since the angle between the velocity (v) and the magnetic field (B) is 90 degrees (perpendicular) sin(θ)=1. ∣F∣=∣q∣⋅∣v∣⋅∣B∣⋅sin(θ)=(1.6×10−19C)⋅(3×10 5m/s)⋅(0.8T)⋅1 Now, calculate the result: F=4.8×10−14 N The direction of the magnetic force is perpendicular to both v and B and follows the right-hand rule. It will be in the positive y-direction.