Video Transcript
A thin, circular coil of wire with
a radius of 2.3 centimeters has 28 turns. The coil carries a constant current
of 330 milliamperes. The strength of the magnetic field
produced is measured to be 𝐵 teslas at the center of the coil. After the measurement of the
magnetic field, the coil is reshaped so that it has the same length but six fewer
turns of wire. The current in the coil is then
adjusted until the strength of the magnetic field produced at the center of the coil
is 𝐵 teslas. Calculate the new value of the
current. Give your answer in milliamperes to
the nearest whole number. Use a value of four 𝜋 times 10 to
the negative seven tesla meters per ampere for 𝜇 naught.
In this question, we’re given a
circular coil of wire. We need to work out how the current
in the wire must change to keep the magnetic field strength at the center of the
coil constant when the number of turns is decreased. Let’s start by clearing some space
and thinking about the initial setup of the coil.
Here, we haven’t drawn all 28 turns
in the coil, but just a representative number of them. We know that the strength of the
magnetic field produced is equal to 𝐵 teslas at the center of the coil. To answer this question then, we’ll
need to use the formula that gives us the magnetic field strength at the center of a
coil of wire. The magnetic field strength, 𝐵, at
the center of a coil of wire with 𝑁 turns is equal to 𝜇 naught 𝑁𝐼 divided by two
𝑟, where 𝐼 is the current in the wire and 𝑟 is the radius of the coil. Since we’re talking about the
coil’s first setup, let’s label all the variables on the right-hand side of this
formula with the subscript one.
We have been given the values of
these quantities. 𝑟 one is 2.3 centimeters, 𝑁 one
is 28 turns, and 𝐼 one is 330 milliamps. Now, we could solve this question
by substituting these values into our formula, finding the value of 𝐵, and then
using this value to work out the current in the second setup of the coil. However, there’s actually a simpler
method, which lets us skip a few calculations. So, let’s take a look at that.
To solve this problem, we need to
think about the second setup of the coil after it has been adjusted. We know that the coil is set up
such that the magnetic field strength at the center is still equal to 𝐵 teslas. We can express the magnetic field
strength 𝐵 using the same formula as before, except now we’ll label our quantities
with the subscript two. 𝐵 equals 𝜇 naught 𝑁 two 𝐼 two
divided by two 𝑟 two. We now have two expressions for 𝐵:
𝐵 equals 𝜇 naught 𝑁 two 𝐼 two divided by two 𝑟 two and 𝐵 equals 𝜇 naught 𝑁
one 𝐼 one divided by two 𝑟 one.
Since we know that 𝐵 has the same
value in both setups of the coil, we can actually equate these two expressions: 𝜇
naught 𝑁 two 𝐼 two divided by two 𝑟 two equals 𝜇 naught 𝑁 one 𝐼 one divided by
two 𝑟 one. Here, we can cancel out the factors
of 𝜇 naught and two. We’re not told that the radius of
the coil changes. So we can also assume that 𝑟 two
is equal to 𝑟 one and cancel these terms too. This leaves us with a much simpler
equation: 𝑁 two times 𝐼 two equals 𝑁 one times 𝐼 one.
In this question, we are asked to
find the current in the second coil, 𝐼 two. We can rearrange this expression to
make 𝐼 two the subject by dividing both sides by 𝑁 two. This leaves us with the formula 𝐼
two is equal to 𝑁 one times 𝐼 one divided by 𝑁 two. Now, all we need to do is
substitute in these values. We know that 𝑁 one is 28 and 𝐼
one is 330 milliamps. We also know that 𝑁 two is six
fewer than 𝑁 one. 28 minus six gets us a value of
22. Substituting these in, we find that
𝐼 one is equal to 28 times 330 milliamps divided by 22. Plugging this into a calculator, we
get a value of 420 milliamps.
So, this is our final answer to
this question: the new current in the adjusted coil must have a value of 420
milliamps.
