Video Transcript
A loop of wire has a
self-inductance of 0.36 henrys. The current through the loop is 420
milliamps. What is the magnetic flux produced
by the current? Give your answer to two decimal
places.
Okay, so in this question, we’ve
got a loop of wire which we can draw like this. We are told that the loop carries a
current of 420 milliamps, which we’ve labeled as 𝐼. Because of this current, a magnetic
field is produced that passes through the loop. And because we have this field
passing through the area of the loop, that means that we’ve got a magnetic flux
through the loop. The value of this magnetic flux is
exactly what the question is asking us to find. The other information that we’re
given is about the self-inductance of the loop of wire. We’ll label this self-inductance as
𝐿, and we’re told that it has a value of 0.36 henrys.
We can recall that the
self-inductance 𝐿 of a loop of wire is equal to the ratio of the magnetic flux 𝜙
subscript 𝑚 produced by a current 𝐼 in that loop to the current itself 𝐼. So we have that the self-inductance
𝐿 is equal to the magnetic flux 𝜙 subscript 𝑚 divided by the current 𝐼. In this equation, if the magnetic
flux has units of weber and the current has units of amperes, then the
self-inductance is in units of henrys. In our case, we know the value of
the self-inductance 𝐿 and we know the current 𝐼 through the loop of wire. We want to solve for the magnetic
flux 𝜙 subscript 𝑚. This means that we need to
rearrange the equation to make 𝜙 subscript 𝑚 the subject.
To do this, we multiply both sides
of the equation by the current 𝐼. On the right-hand side of the
equation, this 𝐼 in the numerator cancels the 𝐼 in the denominator. Then, writing the equation the
other way round, we have that the magnetic flux 𝜙 subscript 𝑚 is equal to the
self-inductance 𝐿 multiplied by the current 𝐼. We know that to get a magnetic flux
in units of weber, we need a self-inductance in henrys and a current in amps. Our value for the self-inductance
𝐿 is indeed in units of henrys. However, our value for the current
𝐼 is in milliamps rather than amps.
To convert from milliamps to amps,
we need to recall that 1000 milliamps is equal to one amp. Or equivalently, one milliamp is
equal to one thousandth of an amp. So to get a current 𝐼 in units of
amps, we take the value in units of milliamps and multiply by one over 1000 amps per
milliamp. The units of milliamps cancel with
the per milliamp, and this leaves us with units of amps. So we have that 𝐼 is equal to 420
divided by 1000 amps. This works out as a value of 0.42
amps.
Now that we’ve got our current in
units of amps, we’re ready to take our values for the self-inductance 𝐿 and the
current 𝐼 and substitute them into this equation to calculate the magnetic flux 𝜙
subscript 𝑚. When we do this, we get that 𝜙
subscript 𝑚 is equal to 0.36 henrys, that’s our value for 𝐿, multiplied by 0.42
amps, which is our value for 𝐼. Because the self-inductance has
units of henrys and the current has units of amps, then the magnetic flux 𝜙
subscript 𝑚 will have units of weber. Evaluating the expression gives a
result of 0.1512 weber. The last thing left to do is to
notice that we were asked to give our answer to two decimal places. Rounding to two decimal places
gives us our final result that the magnetic flux produced by the current is equal to
0.15 weber.
