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.
