Video Transcript
A solenoid is formed of 35 turns of
wire over a length of 42 millimeters. The magnetic field at the center of
the solenoid is measured to be 4.9 times 10 to the negative four tesla. Calculate the current in the
wire. Give your answer in amperes to two
decimal places. Use a value of four 𝜋 times 10 to
the negative seven tesla meters per ampere for 𝜇 naught.
This question is asking us about a
solenoid, which is a wire formed into a shape like this one shown here, consisting
of a series of equally spaced loops or turns. In this case, we’re told that the
solenoid has 35 of these turns. And let’s label this number as
𝑁. We’re also told that these turns of
wire are distributed over our length of 42 millimeters. And let’s label this length as
𝐿. It’s worth being clear that this
length 𝐿 is the distance between these two ends of the solenoid, not the total
length of the wire that’s used to form it.
The other information that we’re
given in the question is the strength of the magnetic field at the center of the
solenoid. We’re told that this is measured to
be equal to 4.9 times 10 to the negative four tesla. And we’ve labeled this magnetic
field as 𝐵. Now, the reason that there’s a
magnetic field inside of the solenoid is that there’s a current, which we’ll label
as 𝐼, in the wire. The value of this current is what
we’re asked to work out in this question.
In order to do this, let’s recall
that there’s an equation which relates the magnetic field 𝐵 inside of a solenoid to
the current 𝐼 in the wire, the number of turns 𝑁 of that wire, and the length 𝐿
of the solenoid. Specifically, 𝐵 is equal to 𝜇
naught multiplied by 𝑁 multiplied by 𝐼 divided by 𝐿, where 𝜇 naught is a
constant known as the permeability of free space.
Since we’re trying to find the
value of the current 𝐼, we need to rearrange the equation in order to make 𝐼 the
subject. To do this, we multiply both sides
of the equation by 𝐿 over 𝜇 naught 𝑁. On the right-hand side of the
equation, we can see that the 𝐿s, the 𝜇 naughts, and the 𝑁s cancel from the
numerator and denominator. Then, if we swap over the left- and
right-hand sides of the equation, we have that the current 𝐼 is equal to 𝐵 times
𝐿 divided by 𝜇 naught times 𝑁.
Before we substitute in our values
to the right-hand side of the equation, we should take a moment to think about the
units of the quantities. We’ve been given a value for the
constant 𝜇 naught with units of tesla meters per ampere. Within this unit, we have length
units of meters. But the length 𝐿 of the solenoid
is given in units of millimeters.
To make these units compatible, we
need to convert the value of 𝐿 from millimeters into meters. To do this, let’s recall that the
unit prefix lowercase m or milli- means a factor of one over 1000. This means that one millimeter is
equal to one thousandth of a meter. And so to convert from units of
millimeters into units of meters, we multiply by a factor of one over 1000, or
equivalently we divide by a factor of 1000. Applying this to the solenoid
length 𝐿 of 42 millimeters, we have that 𝐿 is equal to 42 divided by 1000
meters. This works out as a length of 0.042
meters.
We’re now ready to take our values
for the quantities 𝐵, 𝐿, and 𝑁 along with the value of the constant 𝜇 naught and
substitute them into this equation to calculate the current 𝐼. When we do that, we end up with
this expression here for the current 𝐼. In the numerator, we’ve got 4.9
times 10 to the negative four tesla, which is our value for 𝐵, the strength of the
magnetic field inside of a solenoid, multiplied by 0.042 meters, which is the
solenoid’s length 𝐿. Then, in the denominator, we’ve got
four 𝜋 times 10 to the negative seven tesla meters per ampere. That’s our value for the constant
𝜇 naught. And this is multiplied by 35, which
is the number of turns of wire 𝑁.
We can notice that both the units
of teslas and meters will cancel from the numerator and the denominator of the
fraction. That leaves us with overall units
for the current 𝐼 of one divided by one over amperes, which we can notice is the
same as just units of amperes. Evaluating this expression, we get
a result for 𝐼 of 0.4679 et cetera amperes.
Let’s notice that we’re asked to
give our answer in amperes to two decimal places. Our result for the current 𝐼 is
already in the correct units of amperes. So we just need to round this value
to two decimal places of precision. To two decimal places, the result
rounds up to 0.47 amperes. Our answer then is that for the
solenoid in this question, the current in the wire is equal to 0.47 amperes.
