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

Answer:

$6.6\cdot 10^{-3} N$

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:

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

Answer 2

Answer:

6.6 × 10–3 N

Explanation: