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

Answer:

0.047 T

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:

$B=\frac{F}{IL sin \theta}=\frac{0.70}{(10.0)(1.50)(sin 90^{\circ})}=0.047 T$