Question Video: Determining the Change in the Magnetic Field Strength That Induces an emf in a Conducting Coil Physics • Third Year of Secondary School

Video Transcript

A conducting coil with four turns has a diameter 𝑑 equals 25 centimeters. The coil moves 1.5 centimeters at a velocity 𝑣 equals 7.5 centimeters per second parallel to the axis of a stationary bar magnet, as shown in the diagram. An emf of magnitude 3.6 millivolts is induced in the coil while it moves past the magnet. Find the change in the magnetic field strength between the points where the coil started to move and where it stopped moving. Give your answer in scientific notation to one decimal place.

In our diagram, we see these two components: there’s the conducting coil that has four turns and there’s a stationary permanent magnet. The bar magnet, we know, generates a magnetic field. And as the conducting coil moves along with a velocity 𝑣, it experiences a change in the magnetic field due to the magnet. That change in magnetic field through the area of the conducting coil creates a change in magnetic flux, and that by Faraday’s law induces an emf in the coil. In this example, we want to find the change in the magnetic field strength between the points where the coil started to move and where it stopped. We can begin by writing down some of the information given to us.

We’re told that the coil moves at a velocity of 7.5 centimeters per second. Along with this, the distance the coil moves, we’ll call it 𝑠, is 1.5 centimeters. And this movement through the bar magnet’s magnetic field induces an emf, we’ll call 𝜀, of 3.6 millivolts in the coil. Along with all this, the coil diameter 𝑑, we’re told, is 25 centimeters.

Clearing some space on screen, we write down the law we mentioned earlier, Faraday’s law. This law connects a change in magnetic flux, Δ𝜙 sub 𝐵, through a conductor with an emf, represented by 𝜀, induced in the conductor. In this equation, capital 𝑁 is equal to the number of turns in the conductor, and Δ𝑡 is the time interval over which this change in magnetic flux Δ𝜙 sub 𝐵 occurs. In general, magnetic flux, 𝜙 sub 𝐵, is equal to the strength of a magnetic field, 𝐵, multiplied by the area exposed to that field, 𝐴. This means that as we go to apply Faraday’s law, we can replace 𝜙 sub 𝐵 here with the quantity 𝐵 times 𝐴. Here, 𝐵 is the magnetic field created by our bar magnet and 𝐴 is the area of one of the loops of our conducting coil.

We can see that as our coil moves, the area 𝐴 of those loops doesn’t change. Specifically, the area exposed to the magnetic field doesn’t change. However, because the magnetic field around the bar magnet is not uniform, as the coil moves, the magnetic field strength 𝐵 does change. The fact that in this quantity 𝐵 changes while 𝐴 does not means that we can rewrite the equation like this. It’s this variable Δ𝐵 representing the change in magnetic field experienced by the coil that we want to solve for. Knowing that, let’s rearrange this equation so that Δ𝐵 is the subject. Multiplying both sides of the equation by Δ𝑡 divided by negative 𝑁 times 𝐴 causes negative 𝑁, 𝐴, and Δ𝑡 to cancel out on the right.

If we then switch the sides of the remaining expression, we find that Δ𝐵 equals 𝜀 times Δ𝑡 over negative 𝑁 times 𝐴. In our problem statement, we’re given values for 𝜀, 𝑁, and 𝐴. We didn’t write it down earlier, but 𝑁, the number of turns in our coil, is four. We don’t, however, have a value for Δ𝑡, the time over which our conducting coil is moving. But we do know how far the coil moved, 1.5 centimeters, and we know how fast it moved, 7.5 centimeters per second.

If we recall that, in general, the average speed of an object equals the distance traveled by that object divided by the time taken to travel that distance, then multiplying both sides of this equation by 𝑡 divided by 𝑣 — so that on the left, 𝑣 cancels and on the right, the time 𝑡 cancels — we find that this equation for average speed also says that 𝑡 equals 𝑑 divided by 𝑣. Using the variables in our scenario then, we can write that Δ𝑡 equals 𝑠, the distance traveled by the coil, divided by 𝑣, its speed. Making this substitution in our equation, it then reads that Δ𝐵 equals 𝜀 times 𝑠 divided by negative 𝑁 times 𝐴 times 𝑣.

We’re given values for all of the variables on the right side of this expression, except for the cross-sectional area 𝐴. Note, though, that we do know the diameter of each of the individual loops of our coil. We can remember that the area of a circle, in terms of that circle’s diameter 𝑑, is equal to 𝜋 divided by four times 𝑑 squared. Replacing the area 𝐴 in our equation with this expression, we can note that this is mathematically equivalent to multiplying numerator and denominator by four, giving us this expression.

At this point, let’s note that this negative sign in our equation, which comes from the negative sign in Faraday’s law, helps to indicate the direction of the emf induced in a conductor. In this example, though, we’re only told the magnitude of the emf induced. And therefore, when we calculate Δ𝐵, we’ll only be able to calculate the magnitude of its change. Because of this, we can remove the negative sign from this expression. Our answer will be positive. So we don’t need to consider the direction of this change. We’re now ready to substitute in the known values for 𝜀, 𝑠, 𝑁, 𝑑, and 𝑣. With these substitutions made, notice that none of our units are in SI base units. That is, we’ll want to convert millivolts to volts and centimeters to meters.

Before we do this, notice that there’s a factor of four in numerator and denominator that can therefore cancel out. When it comes to our unit conversions, we recall that one millivolt equals 10 to the negative three or one one thousandth of a volt. This means that 3.6 millivolts equals 3.6 times 10 to the negative three volts. Likewise, one centimeter equals 10 to the negative two or one one hundredth of a meter. And so, we can convert all of our centimeter values to meters by multiplying by 10 to the negative two.

The units in this expression, and feel free to confirm this, work out to volt seconds per meter squared, which is exactly equivalent to teslas. Calculating this result to one decimal place, we find that Δ𝐵 is 3.7 times 10 to the negative three teslas. This is the magnitude of the change in magnetic field experienced by our moving coil.