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?

A) 6.6 × 10–3 N
B) 4.9 × 10–3 N
C) 4.9 × 103 N
D) 6.6 × 103 N

Answer 1

The magnitude of the magnetic force acting on the charge is A) 6.6 × 10–3 N.

Answer 2

A) 6.6 × 10–3 N

Step-by-step explanation:

Given

Charge
`q = 5.7 \mu C = 5.7 * 10^(-6) C`

Velocity
`v = 4.5 * 10^5 m/s`

Field strength
`B = 3.2 mT = 3.2 * 10^(-3) T`

Angle made with vertical
`\phi = 37^o`

Solution

Angle between the field and velocity

`\theta  = 90 - \phi\\\\\theta = 90 - 37\\\\\theta = 53^o`

Force on a charged particle in a magnetic field

`F = qvBsin\theta\\\\F = 5.7 * 10^(-6) * 4.5 * 10^5 * 3.2 * 10^(-3) * sin53\\\\F = 6.6 * 10^(-3) N`