While stringing holiday lights on his home, James pulls the 4.0m string of lights up the side of his home so that the wire is lying perpendicular to EArth's magnetic field creating a magnetic force...

While stringing holiday lights on his home, James pulls the 4.0m string of lights up the side of his home so that the wire is lying perpendicular to EArth's magnetic field creating a magnetic force of 7.39X10^-4N. The cord is carrying a current of 6A. Calculate the Earth's magnetic field at James's location.

 

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This problem can be answered once you understand the relationship between the current in a length of wire and the magnetic field.

The generic relationship is:

`F=BIlsin(theta)` where `B` is the magnetic field strength, `I` is the current in the wire, `l ` is the length of the wire, and `theta` is the angle between the wire and the magnetic field.

Rearrange the equation to solve for B..

`B=F/(Ilsin(theta))`

From here, we plug-in the givens from the problem and determine the answer.

`B=(7.39times10^(-4)text(N))/((6text(A))(4.0text(m))sin(90))~~3.08times10^-5text(T)`

This value falls within the same order of magnitude of typical measurements of Earth's magnetic field.

Sources:

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