Question

What is the magnetic force on a 240 mm length of wire stretched between two towers carrying a 150 A current given that the earth’s magnetic field is 5×10-5T and makes an angle of 60o with the wire?

Answer 1

We will have the following:

First, we recall that:

$F=IBlsin(\theta)$

Now, from the information provided we will have that:

length = 0.24m

Current = 150A

Magnetic field = 5*10^-5T

Angle = 60°

So:

$\begin{gathered} F=(150A)(5\ast10^{-5}T)(0.24m)sin(60)\Rightarrow F=1.558845727\ast10^{-3}N \\ \\ \Rightarrow F\approx1.56\ast10^{-3}N \end{gathered}$

So, the force is approximately 1.56*10^-3 N.

***Explanation***

We know that the magnetic force is given by the expression:

$F=IBlsin(\theta)$

Here "I" is the current flowing, "B" is the magnetic field, "l" is the length of the wire in meters and theta is the angle of the magnetic field with respect to the wire.

When we replace the values we obtain the force of the magnetic field.