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

15 N

Step-by-step 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`