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

Answer:

A) 6.6 × 10–3 N

Explanation:

Given

Charge $q = 5.7 \mu C = 5.7 \times 10^{-6} C$

Velocity $v = 4.5 \times 10^5 m/s$

Field strength $B = 3.2 mT = 3.2 \times 10^{-3} T$

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

$F = qvBsin\theta\\\\F = 5.7 \times 10^{-6} \times 4.5 \times 10^5 \times 3.2 \times 10^{-3} \times sin53\\\\F = 6.6 \times 10^{-3} N$