Question

A magnetic field has a magnitude of 1.2e-3 T, and an electric field has a magnitude of 5.1e3 N/C. Both fields point in the same direction. A positive -1.8 microC moves at a speed of 3.5e6 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

Magnetic field, B = 1.2 x 10^-3 T

Electric field, E = 5.1 x 10³ N/C

Charge, q = 1.8 x 10^-6 C

Velocity, v = 3.5 x 10⁶ m/s

Angle, θ = 90 deg

Force due to magnetic field, Fm acts in a direction perpendicular to the magnetic field and velocity, where as the force due to electric field, Fe acts in a direction parallel to the electric field. So, Fm and Fe

are perpendicular. Force due to magnetic field:

Fm = B q v sinθ

       = 1.2 x 10^-3 * 1.8 x 10^-6 * 3.5x 10⁶ * sin 90

       = 7.5 x 10^-3 N

Force due to electric field:

F_e = E q

     = 5.1 x 10³ * 1.8 x 10^-6

     = 9.18 x 10^-3 N

As Fm and Fe are acting perpendicular,

Net force, F = √ [ F_m² + F_e² ]

                    = √ [ (7.5 x 10^-3)² + (9.18 x 10^-3)² ]

                    = 1.18 x 10^-2

                    = 0.0118 N

F =0.0118 N