Question

A wire that is 1.50 m long and carrying a current of 10.0 A is at right angles to a uniform magnetic field. The force acting on the wire is 0.70 N. What is the strength of the magnetic field?

Answer 1

0.047 T

Step-by-step explanation:

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

`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 B and I

In this problem:

L = 1.50 m is the length of the wire

I = 10.0 A is the current in the wire

F = 0.70 N is the force exerted on the wire

`\theta=90^(\circ)` since the current is at right angle with the field

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

`B=(F)/(IL sin \theta)=(0.70)/((10.0)(1.50)(sin 90^(\circ)))=0.047 T`