The current through a wire that is 0.80 m long is 5.0 A. The wire is at right angles to a 0.60 T magnetic field. What is the magnitude of the force on the wire?

0.24 N

2.4 N

24 N

240 N

Answer: B

F = 2.4 N

Explanation: Given that

Magnetic field B = 0.60 T

Current I = 5 A

Length L = 0.8 m

The force on an individual charge moving at the drift velocity Vd is given by F = qVdB sin θ.

Since current = charge/time

I = q/t

q = It

Also, recall that V = L/t. Therefore

F = I * t * L/t * B * sinØ

F = ILBsin θ

Since the wire is at right angles to the magnetic field

Sin θ = sin 90 = 1 then,

F = 5 * 0.8 * 0.6 = 2.4N

The force on the wire is 2.4 N