Question

What is the force, in units of milliNewtons (mN), exerted on a wire that carries a current of 4.7 A when there is 3.5 cm of the wire in the magnetic field of strength 0.97 T and the wire is perpendicular to the direction of the magnetic field?

Answer 1

160 mN

Step-by-step explanation:

The force exerted on the wire is given by

`F=BIL sin \theta`

where

B is the magnetic field strength

I is the current in the wire

L is the length of the piece of wire

`\theta` is the angle between the direction of the field and the wire

In this problem:

B = 0.97 T

I = 4.7 A

L = 3.5 cm = 0.035 m

`\theta=90^(\circ)`

Therefore, the force exerted on the wire is

`F=(0.97 T)(4.7 A)(0.035 m)(sin 90^(\circ))=0.160 N=160 mN`