# Video: Physics Past Exam • 2017/2018 • Pack 1 • Question 2A

Physics Past Exam • 2017/2018 • Pack 1 • Question 2A

01:36

### Video Transcript

In a transformer, the rate of change of current in the primary coil is five amps per second and the back EMF induced in the secondary coil is four volts. What is the mutual induction coefficient of the coils?

Drawing a sketch of a transformer, we see it consists of a primary coil of wire and a secondary coil connected together by a common core. Electrical current flows into the primary coil. And we’re told the rate of change of that current, five amps per second.

This change in current induces a change in magnetic field within the windings of the primary coil. And these magnetic flux lines are directed by the core through the windings of the secondary coil.

An EMF will be induced in the secondary coil that opposes the changing magnetic field through its coils. This is known as the back EMF, and we’re told its magnitude, four volts.

Given this information, we want to solve for the mutual induction coefficient of the coils, that is, to what extent the coils induce action in one another. If we represent this coefficient with the letter capital 𝑀, then that coefficient multiplied by the change in current divided by the change in time in the primary coil is equal to the EMF that’s induced in the secondary coil.

We can rearrange this equation to solve for the mutual induction coefficient, 𝑀. In our case, the EMF induced is equal to four volts, and Δ𝐼 over Δ𝑡 is equal to five amps every second. This fraction is equal to 0.8 henries. That’s the mutual induction coefficient of the coils.