Question

A wire carries a current of 10 amps in a direction of 90 degrees with respect to the direction of an external magnetic field of strength 0.3 tesla. What is the magnitude of the magnetic force on a 5m length of wire?​

Answer 1

Answer:

15 N

Explanation:

The magnetic force on a piece of current-carrying wire is given by:

$F=ILB sin \theta$

where

I is the current in the wire

L is the length of the piece of wire

B is the magnetic field strength

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

In this problem:

I = 10 A

B = 0.3 T

L = 5 m

$\theta=90^{\circ}$

Substituting into the equation, we find

$F=(10 A)(0.3 T)(5 m) sin 90^{\circ}=15 N$