Video Transcript
Two long straight parallel
conducting wires are separated by distance 𝑑, as shown in the diagram. The wires both carry currents of
1.6 amperes in the same direction. A length 𝐿 equals 0.75 meters of
each wire exerts a force of 3.5 micronewtons on the other. Find the distance 𝑑. Use a value of four 𝜋 times 10 to
the negative seven henries per meter for the magnetic permeability of the region
between the wires. Give your answer to two decimal
places.
In this question, we have two
parallel current-carrying wires, which exert forces on one another that are equal in
magnitude. We know that here each wire exerts
a force of magnitude 3.5 micronewtons. We’ll label this force as 𝐹. Let’s recall the formula for
determining the force, 𝐹, on either current-carrying wire. 𝐹 equals 𝜇 naught times 𝐼 one
times 𝐼 two times 𝐿 divided by two 𝜋𝑑, where 𝜇 naught is the permeability of
free space. 𝐼 one and 𝐼 two give the current
in each wire. 𝐿 is the length of wire we’re
interested in. And 𝑑 is the distance between the
wires.
This question is asking us to solve
for the distance between the wires, so we should rearrange the equation to make 𝑑
the subject. To do this, let’s clear space on
screen and copy the equation and multiply both sides by 𝑑 over 𝐹. This way, 𝑑 cancels out of the
right-hand side and force cancels out of the left-hand side, leaving 𝑑 by
itself. Thus, the equation now reads 𝑑
equals 𝜇 naught times 𝐼 one times 𝐼 two times 𝐿 divided by two 𝜋𝐹.
Before we start calculating, let’s
get organized and write out each of the terms we need to solve this question.
First, we were told that 𝜇 naught,
the permeability of the empty space between the wires, equals four 𝜋 times 10 to
the negative seven henries per meter. We were also told that both wires
carry a current of 1.6 amperes, so 𝐼 one and 𝐼 two both equal 1.6 amps.
Next, although the wires extend
beyond the length 𝐿, we’re only concerned with the wires’ interaction within this
length. So for our calculation, we will
just use the given value of 𝐿 equals 0.75 meters. And finally, we’ve already noted
that the force between the wires is 3.5 micronewtons. Let’s recall that the prefix micro-
means 10 to the negative six. So we should rewrite the force 𝐹
as 3.5 times 10 to the negative six newtons.
We’re now ready to substitute all
these values into the formula. Now, there are a lot of units at
work here, so let’s take a closer look at them before we calculate.
First, we might recall that
henries, the SI derived unit of inductance, are equal to newton meters per amperes
squared. So let’s make this substitution in
the formula. Thus, the units for the
permeability can be written as newton meters per ampere squared per meter squared
or, equally, newton meters per ampere squared meter. So meters in the numerator and
denominator cancel each other out.
Next, notice that we can cancel the
newtons from this term with newtons in the denominator from force. We can also cancel per ampere
squared with amperes from the two current terms. Notice, now, that meters are the
only unit associated with this entire expression. This is a good sign, since after
all we’re solving for a distance.
Finally, let’s go ahead and
calculate. Plugging this into a calculator
gives a result of 0.1097 and so on meters. The last thing we need to do is
round this to two decimal places, so our final answer is 0.11 meters. Thus, we have found that the
distance 𝑑 between the two conducting wires is 0.11 meters.
