Question

A point charge of 5.7 µC moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mT, as shown in the diagram.

What is the magnitude of the magnetic force acting on the charge?

6.6 × 10–3 N
4.9 × 10–3 N
4.9 × 103 N
6.6 × 103 N

Answer 1

6.6 × 10–3 N

Step-by-step explanation:

Answer 2

`6.6\cdot 10^(-3) N`

Step-by-step explanation:

The magnitude of the magnetic force acting on the charge is given by

`F=qvB sin \theta`

where

`q=5.7 \mu C = 5.7 \cdot 10^(-6) C` is the magnitude of the charge

`v=4.5\cdot 10^5 m/s` is the velocity of the charge

`B=3.2 mT = 0.0032 T` is the magnitude of the magnetic field

`\theta = 90^(\circ)-37^(\circ) =53^(\circ)` is the angle between the directions of v and B

Substituting the numbers into the equation, we find:

`F=(5.7\cdot 10^(-6))(4.5\cdot 10^5)(0.0032)(sin 53^(\circ))=6.6\cdot 10^(-3) N`