Question Video: Determining Potential Difference Using Mutual Inductance | Nagwa Question Video: Determining Potential Difference Using Mutual Inductance | Nagwa

# Question Video: Determining Potential Difference Using Mutual Inductance Physics • Third Year of Secondary School

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A transformer with an iron core has a primary coil that has 75 turns and a secondary coil that also has 75 turns. The coils have a mutual inductance of 15 H. The current in the primary coil increases the current in the secondary coil at a rate of 1.25 A/s. What is the potential difference across the coils? Give your answer to one decimal place.

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### Video Transcript

A transformer with an iron core has a primary coil that has 75 turns and a secondary coil that also has 75 turns. The coils have a mutual inductance of 15 henries. The current in the primary coil increases the current in the secondary coil at a rate of 1.25 amperes per second. What is the potential difference across the coils? Give your answer to one decimal place.

Transformers can take on different shapes. And let’s say that ours has a straight iron core like this with the two coils, one primary and one secondary wrapped around it. Our problem statement says that both of these coils have 75 turns. In our sketch, we haven’t shown this many, but this sketch is just to give a general idea of what our setup looks like.

We’re told that these two coils have something called a mutual inductance. In general, a loop of conducting material, like this one, experiences inductance when the total magnetic field passing through that loop, called the magnetic flux, changes. Typically, we represent magnetic flux using this symbol: 𝜙 sub 𝐵. If that flux changes Δ𝜙 sub 𝐵 over some amount of time Δ𝑡, then that means an emf, an electromotive force, also known as a potential difference, will be induced across this loop. That emf will then create a current in the loop.

This is why inductance is so important. By changing the magnetic flux through a loop over some amount of time, we can create current in the loop where there was none before. The change in magnetic field experienced by this conducting loop might be generated by this one. That could happen if there is a change in current in this first loop.

When this happens, when one conducting loop creates a change in magnetic field experienced by a second conducting loop which induces current in that second loop, then we say that there is a mutual inductance between the two loops. That’s what’s going on between our primary and secondary coils. They have a mutual inductance, we’ll call it 𝑀, of 15 henries. And we can note here that the unit of the henry is equal to a weber, that’s the unit of magnetic flux, divided by an ampere, the unit of current. We see then how inductance in henries connects flux in webers to current in amperes.

We’ve seen so far that for two fixed conducting loops, like these in our example, the second loop must experience a change in magnetic field leading to a change in magnetic flux in order for current to be induced in that loop. Therefore, the current in what we can call the primary loop must change over time. If we take the change in current in a conducting loop Δ𝐼 and divide that by the amount of time over which the current changes Δ𝑡, then this fraction is related to the emf, the potential difference, induced in the secondary loop. The change in current over time and the potential difference induced are related by the mutual inductance 𝑀.

In our transformer, we’re told that the current in the primary coil changes at a rate of 1.25 amperes per second. That is equal to Δ𝐼 divided by Δ𝑡. If we multiply this quantity by the mutual inductance between the coils 15 henries, then we get the emf where potential difference induced across the secondary coil. Physically, the magnetic field lines generated by the change in current in our primary coil are transferred by the iron core through the loops of the secondary coil. As we’ve seen, this induces a potential difference in the secondary coil and a current.

In this example, though, it’s just the potential difference, represented by this 𝐸, that we want to solve for. 15 henries times 1.25 amperes per second is 18.75 volts. This would be our final answer, except we’re to give our answer to one decimal place. Rounding our answer then, we get 18.8 volts. This is the potential difference induced across the coils.

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