Video: Finding the Strength of a Uniform Magnetic Field from the Force Experienced by a Current-Carrying Wire

When positioned at 90° to a magnetic field, a wire of length 1 m carrying a current of 4 A experiences a force of 0.2 N. What is the strength of the magnetic field?

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Video Transcript

When positioned at 90 degrees to a magnetic field, a wire of length one meters carrying a current of four amperes experiences a force of 0.2 newtons. What is the strength of the magnetic field?

We can see that this here is our wire. We’re told that it carries a current of four amperes and that it’s one meter long. We’re told that this current-carrying wire is in an external magnetic field, which is oriented at 90 degrees to the wire. Under these conditions, the wire is subjected to a magnetic force we can call 𝐹, which is given as 0.2 newtons.

Knowing all this, we want to solve for the strength of the magnetic field, what we’ve called 𝐵 in our sketch. To figure this out, we can recall a mathematical relationship between wire length, current, magnetic field, and force. The magnetic force on a current-carrying wire is equal to the magnetic field strength that the wire is in multiplied by the length of the wire times the current that’s running through it.

In our case, it’s not the magnetic force we want to solve for, but the magnetic field. And we can do that by rearranging this equation. If we divide both sides by the current times the length of the wire, we get this result. 𝐵 is equal to 𝐹 sub 𝐵 over 𝐼 times 𝐿. Looking at the information in our problem statement, we’re told 𝐹 sub 𝐵. That’s 0.2 newtons. We’re also told the current, 𝐼, in the wire of four amps and the length of the wire of one meters. Our next step then is to substitute these values into this equation. 0.2 newtons divided by four amps times one meter equals 0.05 teslas, where a tesla is the unit of magnetic field. It’s often abbreviated just using a capital T. 0.05 teslas then is the strength of the magnetic field in this situation.

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