Video Transcript
A coil with four turns has a diameter 𝑑 equals 16 centimeters. A bar magnet moves 1.2 centimeters into the coil at an angle of 36 degrees to the axis of the coil in a time of 0.16 seconds. An emf of magnitude 4.1 millivolts is induced in the coil. How much does the average magnetic field strength of the magnet change to produce this emf? Give your answer in milliteslas to one decimal place.
In our diagram, we see this bar magnet, which we’re told moves into the turns of this coil. It does so at an angle 𝜃 to the horizontal, where that angle we’re told is 36 degrees. We want to solve for the change in the average strength of the magnetic field experienced by the coil. This change happens due to the motion of the magnet relative to the coil. We can recall that Faraday’s law tells us how an induced emf, represented here by the Greek letter 𝜀, is related to a change in magnetic flux ΔΦ sub 𝐵. 𝜀 is equal to ΔΦ sub 𝐵 divided by the time Δ𝑡 over which that magnetic flux changes all multiplied by negative the number of turns in the coil where the emf is being induced.
Note that in our scenario, we’re told the number of turns in our coil and we’re also told the magnitude of the emf induced in it. We can write then that 𝑁 equals four and the magnitude of 𝜀 is 4.1 millivolts. Returning to Faraday’s law, we know that in order for some emf to be induced in the coil, there must be a nonzero change in the magnetic flux through it. In general, magnetic flux Φ sub 𝐵 equals magnetic field strength 𝐵 multiplied by the area 𝐴 exposed to that field. As we think about magnetic flux in our scenario, we know that because the bar magnet is in motion relative to the coil, the magnetic field strength 𝐵 experienced by the coil does change. On the other hand, the area 𝐴 exposed to that magnetic field is constant all throughout the motion of the magnet.
Therefore, when we think about the change in magnetic flux experienced by our coil, we can express that overall change as Δ𝐵 multiplied by 𝐴. This is because, as we’ve seen, the area 𝐴 doesn’t change during this process, but the magnetic field strength 𝐵 does. Knowing this, let’s also note the fact that in our scenario, this change in magnetic flux occurs over a time interval of 0.16 seconds. That then is Δ𝑡. Our problem statement tells us that it is Δ𝐵 that we want to solve for. Before we do that, we can calculate the area 𝐴 exposed to the magnetic field. This will be the area of one of the turns of our coil as the bar magnet moves through it. We can recall that we’re told the diameter of each one of these turns; it’s equal to 16 centimeters.
And so recalling that in general the area of a circle in terms of its diameter is 𝜋 divided by four times its diameter squared, we might think that our area 𝐴 in this case is also equal to 𝜋 over four times 𝑑 squared. But we need to be careful to recall that our bar magnet is oriented at an angle 𝜃 to the horizontal. This means that the magnetic field lines from this bar magnet will pass through the turns of our coil at that same angle 𝜃. The actual area then of each one of our turns exposed to the magnetic field is 𝜋 over four times 𝑑 squared times the cos of this angle 𝜃. This cos 𝜃 factor accounts for the fact that our magnetic field lines are not perpendicular to the turns of the coil. With all this settled, we’re just about ready to solve for Δ𝐵. Let’s clear some space on screen to do this.
Our particular application of Faraday’s law looks like this. We’re considering the magnitude of the emf induced. And for that reason, we don’t have the negative sign that typically appears on this side of the equation. To rearrange this equation so that Δ𝐵 is the subject, let’s multiply both sides by Δ𝑡 divided by 𝑁 times 𝐴. That way, the factors of 𝑁, 𝐴, and Δ𝑡 all cancel from the right. And if we then switch the sides of the remaining equation, we find that the change in the magnetic field strength Δ𝐵 equals Δ𝑡 times the magnitude of 𝜀 all divided by 𝑁 times 𝐴. Substituting in our known values, we know that Δ𝑡 is 0.16 seconds. The magnitude of 𝜀 is 4.1 millivolts. And the number of turns in our coil is four. The diameter of each of these turns is 16 centimeters. And the bar magnet is oriented at an angle of 36 degrees to the horizontal.
Before we calculate Δ𝐵, we’ll want to change a few units in this expression. In the numerator, we’d like to convert millivolts into volts and in our denominator, centimeters to meters. We can recall the conversion that one millivolt equals 10 to the negative three or one one thousandth of a volt. That tells us that 4.1 millivolts is equal to 4.1 times 10 to the negative three volts. In a similar way, one centimeter equals 10 to the negative two or one one hundredth of a meter so that 16 centimeters is equal to 16 times 10 to the negative two meters. In our denominator, notice that we have a factor of four and one-fourth that will cancel one another out. We can use the resulting expression then to calculate Δ𝐵. Calculating this expression, we get a result of 0.010082 and so on teslas.
Let’s recall though that we want to report our final answer in units of milliteslas rounded to one decimal place. Just like the relationship between a millivolt and a volt, there are 1000 milliteslas in one tesla. That means we’ll take our decimal place and we’ll move it one, two, three spots to the right so that our answer is now 10.082 and so on milliteslas. And now we’ll round this answer to one decimal place. Since the digit after the first decimal place is greater than or equal to five, we’ll round up so that our final answer is 10.1 milliteslas. This is the change in the average strength of the magnetic field experienced by the turns in the coil.
