Video Transcript
A conducting coil with a radius of 2.5 centimeters has 150 turns. The coil moves perpendicularly to a magnetic field that becomes stronger at the rate of 1.8 milliteslas per second. Find the magnitude of the electromotive force induced in the coil. Give your answer in millivolts to two decimal places.
Let’s say that this is our coil that actually has many more turns than the number we’ve shown here. This coil, we’re told, moves perpendicularly to a magnetic field, and that field actually gets stronger over time. The rate at which the field strength changes, we’ll call this Δ𝐵 divided by Δ𝑡, is positive 1.8 milliteslas per second. For every second of time that passes then, the field increases in magnitude by 1.8 milliteslas. This change in magnetic field strength through the turns of the coil induces an emf in the coil. This happens according to Faraday’s law. That law says that the emf induced in some conductor, we’ll represent it using the Greek letter ε, is equal to negative the number of turns 𝑁 in the conductor times the change in magnetic flux through the conductor ΔΦ sub 𝐵 over the time Δ𝑡 during which that change in flux occurs.
Let’s also note that magnetic flux Φ sub 𝐵 by itself is equal to magnetic field strength 𝐵 multiplied by the area 𝐴 exposed to that field. This means we can write ΔΦ sub 𝐵 that appears in Faraday’s law as Δ the quantity 𝐵 times 𝐴. And then in our particular scenario, we note that the cross-sectional area exposed to our magnetic field is not changing. The field is always perpendicular to the turns of the coil. On the other hand, as we’ve seen, the magnetic field strength 𝐵 does change over time. Because 𝐵 changes but 𝐴 does not in our scenario, we can rewrite Δ the quantity 𝐵 times 𝐴 as Δ𝐵 times 𝐴.
Let’s recall at this point that it’s the magnitude of the emf induced in the coil that we want to solve for. Applying Faraday’s law, we can write that the magnitude of ε, the magnitude of the emf induced, equals the number of turns in the coil times ΔΦ sub 𝐵 over Δ𝑡, where ΔΦ sub 𝐵, we’ve seen, can be written as Δ𝐵 times 𝐴. Notice that now on the right-hand side of this expression, we have Δ𝐵 divided by Δ𝑡. That, we’ve seen, is equal to 1.8 milliteslas per second. We also know from our problem statement that the number of turns in our coil 𝑁 is 150. Lastly, we want to know the cross-sectional area 𝐴 in one turn of our coil that’s exposed to the magnetic field.
Because the coil is oriented perpendicularly to the magnetic field, that area will simply equal the area of one of the turns of our coil. We recall that, in general, the area of a circle is equal to 𝜋 times the radius of that circle squared. Our coil, we know, has a radius of 2.5 centimeters. Before we calculate the magnitude of ε, let’s consider the units in this expression. It’s standard practice before making a calculation to convert units into SI base units where possible. That would mean converting milliteslas into teslas and centimeters into meters.
Let’s recall though that in this particular situation, we want to give our answer in units of millivolts. This means that if we don’t convert our magnetic field strength but instead leave it in units of milliteslas while still converting our units of centimeters into meters, then the number we calculate for our final answer will already be in units of millivolts. So then, leaving our units of milliteslas as they are, we recall that one centimeter equals 10 to the negative two or one one hundredth of a meter. Therefore, 2.5 centimeters equals 2.5 times 10 to the negative two meters. And by the way, this is also equal to 0.025 meters.
If we now move ahead with our calculation for the magnitude of ε, as we’ve said, we’ll get a result in units of millivolts. When we calculate that result and round it to two decimal places, it comes out to 0.53 millivolts. This is the magnitude of the electromotive force induced in the coil.
