Question

Crossed E and B Fields. A particle with initial velocity V = (5.85 * 10^3 m/s) j enters a region of uniform electric and magnetic fields. The magnetic field in the region is B= -(1.350 T) k . Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) -0.320 nC. You can ignore the weight of the particle.

Answer 1

Magnetic force is given as

F = q*(vxB)

Electric field = F/q = q*(vxB)/q

E = vxB

E = 5.85*10^3*1.35 = 7.9*10^3 N/C

a)

Magnitude of electric field = E = 7.9*10^3 N/C

and direction is positive x-direction.

b)

E = vxB

E = 7.9*10^3 N/C

and direction is positive x-direction.