# Question Video: Calculating Magnetic Flux Change in a Transformer Physics • 9th Grade

A transformer with an iron core has a primary coil that has 15 turns and a secondary coil that also has 15 turns. In a time of 0.25 s, a potential difference of 12 V is applied across the primary coil, inducing a potential difference of 12 V across the secondary coil. What is the change of the magnetic flux through the iron core?

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### Video Transcript

A transformer with an iron core has a primary coil that has 15 turns and a secondary coil that also has 15 turns. In a time of 0.25 seconds, a potential difference of 12 volts is applied across the primary coil, inducing a potential difference of 12 volts across the secondary coil. What is the change of the magnetic flux through the iron core?

Let’s say that this is our iron core. Here’s the primary coil of the transformer wrapped around that core, and here’s the secondary coil. Notice that we haven’t drawn these coils, having the correct number of turns 15 in each case. But rather than sketching in all of the turns on each coil, we’ll use this sketch just to give an idea of what our system looks like. Let’s say that the coil on the left is the primary coil and that on the right is the secondary coil.

We’re told that a potential difference of 12 volts is applied across the primary coil and that this induces an identical potential difference across the secondary coil. When 12 volts is applied across the primary coil, that creates a current in that coil. That current induces a magnetic field. The magnetic field is channeled by the iron core so that it passes through the turns of the secondary coil. This is how a potential difference is induced across that secondary coil. Specifically, it happens due to a change in the magnetic flux through the turns of the secondary coil in this transformer. We can write a symbol for magnetic flux this way; it’s the Greek letter 𝜙 with a subscript 𝐵.

In this example, we want to calculate a change in magnetic flux; we’ll represent that as Δ𝜙 sub 𝐵. If we take an up-close look at just one of the turns in our secondary coil, we know that this turn will experience some amount of change in magnetic flux. That is, the overall strength of the magnetic field passing through this loop will change over time. It’s that change we’ve seen that induces a potential difference. The potential difference induced across this entire secondary coil is due to the compounding effects of the change in magnetic flux through every single one of the turns on this coil.

If we let Δ𝜙 sub 𝐵 represent the change in magnetic flux through a single turn of our coil, then this change in magnetic flux over some amount of time Δ𝑡 must be multiplied by the total number of turns in our coil, we’ll call this capital 𝑁, in order to give the EMF or equivalently the potential difference established across the secondary coil. Now, in our case, we know the potential difference established across the coil, 12 volts.

In this equation, what we actually want to solve for in this example is Δ𝜙 sub 𝐵. Starting with this general equation then, we can rearrange it by multiplying both sides by Δ𝑡 divided by capital 𝑁. Doing this will mean that on the right-hand side of our expression, capital 𝑁 cancels out as does Δ𝑡. This gives us a relationship where Δ𝜙 sub 𝐵 is the subject.

Considering the values on the right-hand side, we know that Δ𝑡 is 0.25 seconds. 𝜀, which represents the potential difference induced across the secondary coil, is 12 volts. And then capital 𝑁, the number of turns in the secondary coil, is 15. When we calculate this fraction, we’ll get the change in magnetic flux through a single turn of our secondary coil. This is equal to the change in magnetic flux through the iron core. This fraction is equal to 0.2 webers. The weber we recall is the SI unit of magnetic flux. The change in the magnetic flux through the iron core is 0.2 webers.