A magnetic field has a magnitude of 1.20E3 T, and an electric field has a magnitude of 4.70E3 N/C. Both?

Question

A magnetic field has a magnitude of 1.20E-3 T, and an electric field has a magnitude of 4.70E3 N/C. Both fields point in the same direction. A positive 1.8 µC charge moves at a speed of 3.40E6 m/s in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

Answer 1

The force from the electric field is E*q, the force from the magnetic field is q(v X B). These forces will be at right angles to each other; the electric force in the direction of the electric field, and the magnetic force perpedicular to both the B field and the velocity. Since the v and B are at right angles, the magnitude of the cross product is just q*v*B

Electric force Fe = 4.7*10^3 * 1.8*10^-6 = 0.00846 N

Magnetic force = 1.8*10^-6 * 3.4*10^6 * 1.2*10^-3 = 0.007344 N

The resultant force is the sqr root of the sum of the squares of these:

F = 0.0112 N