Video Transcript
Consider two straight parallel
current-carrying wires and a small compass between the wires. One wire is at a distance of 30
centimeters from the compass and is carrying a current of intensity three amps from
below the plane of the compass to above the plane of the compass. The other wire is at a distance of
50 centimeters from the compass, and the compass needle is initially pointing at
this wire. What should the intensity of the
current passing through the second wire be to make the needle of the compass
undeflected? (A) Five amps, (B) 1.8 amps, (C)
8.3 amps, or (D) 1.1 amps.
Let’s start by drawing a diagram of
the setup described. So, we have two straight parallel
wires. And we’re told there’s a small
compass in a certain position between them. We’re also told that the closest
wire to the compass is carrying a current of three amps from below the plane of the
compass to above the plane of the compass. Let’s call this current 𝐼 one. We’re also told that the other wire
carries a current although we’re not told the size or direction of this current. For now then, let’s just call it 𝐼
two.
Now, because the wires go from
below the plane of the compass to above it, that means that in this diagram, we’re
looking at the side of the compass. Let’s change our view so we’re
looking at the scene from above. Here, we’ve drawn the compass a bit
bigger so we can see the needle. And we’re now looking at the cross
section of each wire. Now, in the question, we’re told
that the compass needle is initially pointing toward the furthest away wire, so to
the right in our diagram. And we need to calculate the
intensity of the current in this wire, 𝐼 two, that would prevent the compass needle
from deflecting from this position.
So to answer this question, we’ll
need to recall that a current will create a magnetic field around itself. Let’s think about the magnetic
field produced by the three-amp current, 𝐼 one, in the wire on the left. Magnetic field lines follow a
circular pattern around a current-carrying wire, like this. This means that at the center of
the compass, the direction of the magnetic field produced by 𝐼 one will be this
way. Let’s label the magnitude of the
field at this point 𝐵 one. Now, the needle on a compass will
naturally align itself with the magnetic field it experiences. So, if 𝐵 one was the only magnetic
field here, we’d expect to see the compass needle deflect to this position. However, 𝐵 one is not the only
magnetic field in this scenario. We also have a magnetic field
produced by the current 𝐼 two in the other wire.
Now, the only way the needle on the
compass can stay undeflected from this position is if the magnetic field produced by
𝐼 one and the magnetic field produced by 𝐼 two perfectly cancel each other out at
the position of the compass. In other words, if the magnetic
field strength produced by 𝐼 two at the position of the compass, let’s call this 𝐵
two, is exactly the same magnitude but acting in the opposite direction to 𝐵 one,
then the two fields will cancel out so that the net field strength at the position
of the compass is zero. And if the magnetic field strength
is zero, then the needle won’t deflect. In other words, we require that 𝐵
one, the magnetic field strength at the position of the compass due to 𝐼 one, plus
𝐵 two, the magnetic field strength at the position of the compass due to 𝐼 two, is
equal to zero. We can rearrange this equation to
𝐵 two equals negative 𝐵 one.
Now, at this point, we can recall
that the magnetic field strength measured at a certain distance from a
current-carrying wire is given by the equation 𝐵 equals 𝜇 naught 𝐼 over two 𝜋𝑟,
where 𝜇 naught is the permeability of free space, 𝐼 is the current in the wire,
and 𝑟 is the distance from the wire at which the magnetic field strength is being
calculated. We can use this expression to
rewrite 𝐵 two and 𝐵 one in the equation we’ve written on the right. So, 𝐵 two, that’s the magnetic
field strength at the position of the compass due to 𝐼 two, can be written 𝜇
naught times 𝐼 two over two 𝜋𝑟 two. And 𝐵 one can be written 𝜇 naught
𝐼 one over two 𝜋𝑟 one. Here, 𝑟 two is the distance
between the center of the compass and the wire carrying 𝐼 two. And 𝑟 one is the distance between
the center of the compass and the wire carrying 𝐼 one.
So now all we need to do to find
the answer to the question is solve this equation for 𝐼 two. We can start by noticing that both
sides of the equation contain a factor of 𝜇 naught over two 𝜋. So, we can divide both sides of the
equation by 𝜇 naught over two 𝜋, which leaves us with 𝐼 two over 𝑟 two equals
negative 𝐼 one over 𝑟 one. Next, we can multiply both sides of
the equation by 𝑟 two to give us 𝐼 two equals negative 𝐼 one over 𝑟 one times 𝑟
two. And now, all we need to do is
substitute in each of the known values.
First, we’re told that 𝐼 one, the
current in the wire nearest to the compass, is three amps. Next, we know that 𝑟 one, the
distance of the compass from the nearest wire, is 30 centimeters. And in SI units, this is 0.3
meters. Finally, 𝑟 two, that’s the
distance of the compass from the wire carrying 𝐼 two, is 50 centimeters. And in SI units, that’s 0.5
meters. Evaluating this expression, we get
a value of negative five amps.
Now, we’ve obtained a negative
value here. But all this means is that the
current 𝐼 two must be going in the opposite direction to the positive current 𝐼
one. However, we don’t need to include
this negative sign in our answer, since the question only asks for the intensity, or
magnitude, of the current. That means our final answer to the
question is option (A). The intensity of the current
passing through the second wire to make the needle of the compass undeflected should
be five amps.
