Video Transcript
Three long, straight, parallel,
conducting wires W one, W two, and W three carry currents of 1.6 amps, 1.1 amps, and
negative 1.9 amps, respectively. W one is 2.5 centimeters from W two
and 5.1 centimeters from W three. W two is located between the other
two wires. Find the magnitude of the force per
meter of length on W one perpendicular to W two. Use a value of four 𝜋 times 10 to
the power of negative seven henries per meter for the magnetic permeability of the
region between the wires. Give your answer in scientific
notation to one decimal place.
So, in this question, we have three
long, straight, parallel, conducting wires called W one, W two, and W three. We’re told the distances between
the wires. So we can draw a diagram like
this. We’re also told that each of the
wires carries a current, so let’s show this in our diagram. The wire W one carries a current of
1.6 amps. We’ll call this 𝐼 one. The wire W two carries a current of
1.1 amps, and we’ll call this current 𝐼 two. Finally, the wire W three carries a
current of negative 1.9 amps, and we’ll call this current 𝐼 three. Note that this current has been
given as a negative number to tell us that 𝐼 three acts in the opposite direction
to 𝐼 one and 𝐼 two.
So we’re being asked to work out
the force per unit meter acting on the wire W one. The very first thing we need to
recall is that a current in a wire produces a magnetic field around the wire. And when a wire carries a current
through a magnetic field, it experiences a force depending on the direction of the
current relative to the field. This means that W one will
experience a force due to the magnetic fields produced by both W two and W
three.
So, first, let’s consider the force
acting on W one due to the current 𝐼 two. We can use the right-hand grip rule
to determine the direction of the magnetic field produced by 𝐼 two. The right-hand grip rule states
that if we point the thumb of our right hand in the direction of the current, then
our fingers curl closed in the direction of the magnetic field. Applying this to 𝐼 two, we see
that the magnetic field produced by this current wraps around W two like this. This means that the direction of
the magnetic field at W one is pointing out of the page. And we can represent this on our
diagram like this.
Recall that a current-carrying wire
in a magnetic field will experience a force. This means that W one will
experience a force due to the fact that it’s carrying a current in the magnetic
field produced by W two. We can calculate the direction of
this force using Fleming’s left-hand rule.
To do this, we point the first
finger of our left hand in the direction of the magnetic field 𝐵 two at W one. That’s out of the screen. We then point our middle finger in
the direction of the current in W one, which is to the right. Having done this, then our thumb
will point in the direction of the magnetic force acting on W one, as a result of
the magnetic field produced by the current in W two. In this case, that’s downward. Let’s call this force 𝐹 two, and
we’ll mark it on our diagram.
So now we’ve worked out the
direction of the force that acts on W one due to the current in W two, we can repeat
the process for the force on W one due to the current in W three. Using the right-hand grip rule,
pointing our right thumb to the left in the direction of 𝐼 three, we find that the
magnetic field produced by 𝐼 three wraps around the wire W three like this. This means that the direction of
this magnetic field at W one is into the screen. And we can mark this on our diagram
like this.
Now, we can use Fleming’s left-hand
rule again to show us the direction of the force acting on W one due to the magnetic
field produced by W three. So we point the first finger of our
left hand into the screen. That’s in the direction of 𝐵 three
at W one. Then, we point our middle finger in
the direction of the current 𝐼 one. Now, our thumb points in the
direction of the force acting on W one due to the magnetic field produced by W
three. Let’s call this force 𝐹 three and
mark it on our diagram as well.
Okay, so we can now see that the
two forces acting on W one are in opposite directions. We want to find the net force per
meter acting on W one. Recall that for two parallel wires
carrying currents 𝐼 one and 𝐼 two, the force per unit length on the wire carrying
𝐼 one is given by this expression, where 𝑑 is the distance between the two wires
and 𝜇 naught is the magnetic permeability of the region between them.
Since we have two wires producing
magnetic fields, which both exert forces on W one, the net force per unit length on
W one per meter is given by the sum of both forces. This is given by this expression,
where 𝑑 two is the distance between W one and W two and 𝑑 three is the distance
between W one and W three.
Now, we can simplify this
expression by noticing that each term has a factor of 𝜇 naught times 𝐼 one over
two 𝜋. This enables us to factorize the
expression like this. Now, all we need to do is
substitute in the necessary values. Let’s do this at the bottom of the
screen.
So, first, we need the value of 𝜇
naught. That’s the magnetic permeability
between the two wires. This is given in the question. It’s four 𝜋 times 10 to the power
of negative seven henries per meter. This value is then multiplied by 𝐼
one, which is the current in the wire W one. We’ve already labeled this current
in our diagram, and it has a magnitude of 1.6 amps. The product of these two values is
then divided by two 𝜋. And this expression is multiplied
by 𝐼 two over 𝑑 two plus 𝐼 three over 𝑑 three.
The value of 𝐼 two, that’s the
current in wire W two, is 1.1 amps. The value of 𝑑 two, the distance
between the wires W one and W two, is 2.5 centimeters, which in SI units is 0.025
meters. 𝐼 three, the current in wire W
three, is negative 1.9 amps. And finally 𝑑 three, the distance
between wires W one and W three, is 5.1 centimeters, or in SI units 0.051
meters.
Note that all of the values we’ve
substituted into this expression are all expressed in units derived from the SI base
units. So we know that our answer will
also be expressed in SI-derived units. Since we’re calculating a force per
unit length, this means our answer will be expressed in units of newtons per
meter.
Now, using a calculator to evaluate
this expression, we obtain a result of 2.158 and so on times 10 to the power of
negative six newtons per meter. The only thing left to do now is
round our answer to one decimal place. This gives us a final answer of 2.2
times 10 to the power of negative six newtons per meter.
So, if we have three parallel wires
each carrying a current as described in the question, the resulting magnetic fields
will produce a force per meter acting on wire W one equal to 2.2 times 10 to the
power of negative six newtons per meter.
