Question

A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire is 0.40 N. What is the strength of the magnetic field?

Answer 1

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

$\theta$ is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

$\theta=90^{\circ}$ since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

$B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T$