### Video Transcript

A transformer consists of a primary
coil and a secondary coil, each with the same number of turns, wrapped around an
iron core. The coils have a mutual inductance
of 32 millihenries. A current in the primary coil
increases the magnetic flux through the core by 4.48 milliwebers. The current induced in the
secondary coil is 1.4 amperes. How many turns does the coil
have?

Let’s say that this is our
transformer, with the primary coil on the left and the secondary coil on the
right. We’re told that these two coils
have a nonzero mutual inductance. This means, for example, that if
the current in the primary coil changes — we’ll refer to this change as Δ𝐼 sub one
— then multiplying this change by the mutual inductance between the coils — we’ll
call this value 𝑀 — will give a value equal to the total change in magnetic flux
through the secondary coil.

Here, ΔΦ sub 𝐵 is the change in
magnetic flux through the iron core, and 𝑁 two is the number of turns in the
secondary coil. We can think then of ΔΦ sub 𝐵 as
the change in magnetic flux experienced by each turn in the secondary coil, and that
coil, we’re saying, has a total of 𝑁 two turns. Therefore, this product equals the
overall magnetic flux through the secondary coil. And it’s this value 𝑁 two that we
want to solve for. Notice though that, physically,
it‘s the change in current in the other coil, in the primary coil, that leads to
this change in magnetic flux through the secondary coil.

Interestingly, mutual inductance 𝑀
works the same way in both directions, we could say. That is, a given change in current
in the primary coil induces an overall change in magnetic flux in the secondary coil
that is equal to the overall change in magnetic flux in the primary coil due to an
equal change in current in the secondary coil. Mathematically, this means we can
write Δ𝐼 sub two, a change in current in the secondary coil, multiplied by the
mutual inductance 𝑀 equals ΔΦ sub 𝐵, the change in flux through the iron core,
times 𝑁 one, the number of turns in the primary coil.

Notice that in these two equations,
the mutual inductance 𝑀 and the change in magnetic flux through the iron core ΔΦ
sub 𝐵 are the same. And actually, we can now recall
that in our problem statement, we’re told the two coils have the same number of
turns. Therefore, 𝑁 one is equal to 𝑁
two. If all that is true, if these three
analogous factors in these two equations are equal to one another, that implies that
the changes in current, Δ𝐼 sub one and Δ𝐼 sub two, must be equal.

This fact is very helpful to us,
because as we’ve seen, it’s 𝑁 two, the number of turns in the secondary coil, that
we want to solve for. Referring to this first equation,
we don’t know Δ𝐼 sub one, the change in current in the primary coil. However, we are told the change in
current in the secondary coil. It’s 1.4 amperes. Therefore, we do know Δ𝐼 sub two,
and by this equation here, these two changes in current are equal. This means we can replace Δ𝐼 sub
one in our first equation with Δ𝐼 sub two. If we do that and then clear some
space below this equation, we now have an equation where we know all the factors
involved except for the one we want to solve for, 𝑁 two.

Dividing both sides of this
equation by ΔΦ sub 𝐵, the change in flux through the iron core, that factor cancels
on the right. And this gives us an expression
that shows that 𝑁 two equals Δ𝐼 sub two multiplied by 𝑀 divided by ΔΦ sub 𝐵. The change in current in the
secondary coil is 1.4 amperes. The mutual inductance 𝑀 is 32
millihenries, and the change in magnetic flux through the core is 4.48
milliwebers.

Before we calculate 𝑁 two, note
that in both the numerator and the denominator, we have a milli- prefix before one
of our units. This prefix indicates a multiplying
factor of 10 to the negative three, or one one-thousandth. That is, 32 millihenries is equal
to 32 times 10 to the negative three henries. Likewise, 4.48 milliwebers is equal
to 4.48 times 10 to the negative three webers. Regarding the units that remain,
the SI unit of inductance, the henry, is defined as a weber, the SI unit of magnetic
flux, divided by an ampere.

If we make this substitution,
replacing henries with webers per ampere, notice then that the factor of amperes
cancels out in our numerator and the factor of webers cancels from numerator and
denominator. Our final answer will be unitless,
as it should be, because we’re calculating a number of turns. Entering this expression on our
calculator, we get a result of exactly 10. This is how many turns this
secondary coil in this transformer has.