Question

A magnetic field has a magnitude of 1.2 × 10−3 T, and an electric field has a magnitude of 4.6 × 103 N/C. Both fields point in the same direction. A positive 1.8 μC charge moves at a speed of 3.1 × 106 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=6.6mN

Explanation:

Given that

Magnetic field=$1.2\times 10^{-3}$ T

Electric field=$4.6\times 10^{3}$ N/C

We know that  when electric field is perpendicular to the motion of the charge so the force on the charge due to electric field will be zero.

Here given that charge is moving perpendicular to the field so force will be zero on the charge.

Here only magnetic force will be act on the charge.

We know that magnetic field given as

F=qvB

Now by putting the value in above formula

$F=1.8\times 10^{-6}\times 3.1\times 10^{6}\times1.2\times10^{-3}$

F=6.6mN