Question

A magnetic field has a magnitude of 1.2 \times 10^{-3} T, and an electric field has a magnitude of 4.6 \times 10^{3}N/C. Both fields point in the same direction. A positive 1.8 \mu C charge moves at a speed of 3.1 \times 10^6 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

Answer: F = 113.4.$10^{-3}$N

Explanation: Net Force is the total forces acting in an object. In this case, there are two forces acting on the charge: one due to magnetic field (Fm) and another due to electric field (Fe). So, net force is

F = Fe + Fm

Force due to electric field

To determine this force:

Fe = q.E, where q is the charge and E is electric field.

Calculating:

Fe = q.E

Fe = 1.8.$10^{-6}$.4.6.$10^{3}$

Fe = 8.28.$10^{-3}$N

Force due to magnetic field: It can only happens when the charge is in movement, so

Fm = q.(v×B), where v represents velocity and B is magnetic field

The cross product indicates that force is perpendicular to the velocity and the field.

Calculating:

Fm = q.v.B.senθ

As θ=90°,

Fm = q.v.B

Fm =  1.8.$10^{-6}$.3.1.$10^{6}$.1.2.$10^{-3}$

Fm = 6.696.$10^{-3}$N

F, Fm and Fe make a triangle. So, using Pythagorean theorem:

F = $\sqrt{Fe^{2} + Fm^{2} }$

F = $\sqrt{(8.28.10^{-3} )^{2} +(6.696.10^{-3} )^{2} }$

F = 113.4.$10^{-3}$N

The net force acting on the charge is F = 113.4.$10^{-3}$N