Video Transcript
In this video, we will learn how to
determine the magnetic forces between parallel lines of current and analyze the net
magnetic fields of the currents. Before we do, we must refresh our
memory about the magnetic field around a single line of current. In order to understand how to find
the net magnetic field around more than one current-carrying wire, we must first
understand how to draw in the magnetic field around a single current-carrying wire
by doing the right-hand rule.
We need to remember that to use the
right-hand rule to determine the direction of magnetic field around a
current-carrying wire that our thumb will point in the direction of the current and
that our fingers will curl around the wire, pointing in the direction of the
magnetic field lines. If we orient a wire such that the
current is directed to the right of our screen, then we must position our hand so
our thumb is also pointing to the right of our screen. And our fingers will curl around
the wire such that magnetic field will be pointing to the bottom of the screen in
front of our wire.
This tells us that the magnetic
field around our current-carrying wire will create circles all along it, represented
by the yellow dotted lines, with the magnetic field being directed into the screen
underneath the wire, out of the screen above the wire, and pointing towards the
bottom of the screen in front of the wire. If we were to look at our wire in a
different direction such that the current was pointing into our screen, applying our
right-hand rule with our thumb being directed into the screen and our fingers
curling around the wire such that the magnetic fields make concentric circles that
are clockwise around our current-carrying wire.
Now that we’ve refreshed our memory
on how to find the magnetic field around a current-carrying wire, let’s determine
how to find the net magnetic field around more than one wire. To determine the net magnetic
field, we need to do vector addition based on the direction of the magnetic field
due to each current-carrying wire at the chosen location. We need to do vector addition
because our magnetic field is a vector. This means that the two magnetic
fields due to the current-carrying wires with magnitudes 𝐵 one and 𝐵 two point in
the same direction. We can add the magnitudes
together. 𝐵 one plus 𝐵 two would be equal
to our net magnetic field.
If, however, our magnetic fields
point in opposite directions, then we must subtract their magnitudes. 𝐵 one minus 𝐵 two will be equal
to the net magnetic fields. Recall from earlier when I said
that a current-carrying wire directed into the screen will have a magnetic field
with concentric circles that are directed clockwise around the wire. Now, let’s set up an identical
current-carrying wire right next to it.
We have drawn a second
current-carrying wire to the right of the first current-carrying wire and
represented the magnetic field with pink dotted lines. If we draw a horizontal line that
goes through both the current-carrying wires, we can then analyze positions where
the magnetic fields overlap to determine whether they are a larger net magnetic
field or a smaller net magnetic field. Let’s go through the labeled
positions below, one, two, and three, and determine whether we should add or
subtract our fields.
At position one, both the yellow
magnetic field and the pink magnetic field are pointing directly to the top of our
screen. Therefore, they’re in the same
direction, which means that we will add the magnitudes of our fields. At position two, the magnetic field
in yellow is pointing to the bottom of the screen and the magnetic field in pink is
pointing to the top of our screen, which means they’re pointing in opposite
directions. So we should subtract the
magnitudes. And finally, at position three,
both the yellow and pink magnetic fields will be pointing to the bottom of the
screen, which means they’re pointing in the same direction. So we will once again add the
magnitudes.
We can see that positions one and
three will have larger net magnetic fields than position two as we added the
individual fields at one and three. To take it one step further, we can
even say that the net magnetic field at position three is the largest. This is because position three is
closer to the wire on the right side than position one is to the wire on the left
side. And position three is also closer
to the wire on the left side than position one is to the wire on the right side.
If we remember that the magnetic
field around a current-carrying wire 𝐵 is equal to the magnetic permeability 𝜇
times the current in the wire 𝐼 divided by two 𝜋 times the distance from the wire
𝑟, then we would recognize that the position that’s closest to the wire or has the
smallest 𝑟 will have the greatest magnetic fields. Of the three positions, position
three will have the largest net magnetic field, then position one, and finally
position two will have the smallest.
Now, let’s look at the force per
unit length between two parallel conducting wires. To determine the force per unit
length, we need to recall two equations. First, the magnetic field around a
current-carrying wire, 𝐵, is equal to the magnetic permeability, 𝜇 naught, times
the current in the wire, 𝐼, divided by two 𝜋𝑟, where 𝑟 is the distance from the
wire. The second equation is that the
force on a current-carrying wire 𝐹 is equal to the current within the wire, 𝐼,
times the length of the wire, 𝐿, times an external magnetic field, 𝐵, that the
wire is placed in.
To derive our expression for force
per unit length, let’s draw on an example of two wires. Our first wire will have a current
𝐼 one directed to the right of our screen, with a length of 𝐿. Using the right-hand rule, where
our thumb will be directed to the right of our screen and our fingers will curl
around our wire, we will see the magnetic field goes into the page on the bottom of
the wire and comes out of the page on the top of the wire.
We then place a second wire into
the field with a current of 𝐼 two directed to the right of the screen and a length
of 𝐿. If the distance between the two
wires is 𝑑, let’s calculate the force exerted on the second wire by the magnetic
field generated by the first wire. We can begin with the equation for
force on our wire. To get the equation in the right
form, we need to divide both sides by 𝐿. Remember, the 𝐼 is the current in
the wire that we’re looking for the force per unit length on. In this case, that would be 𝐼
two. The magnetic field which applies a
force to our second wire comes from current-carrying wire one. And we can find the strength of
that by using our equation for the magnetic field around a current-carrying
wire. The current is current 𝐼 one, and
the distance between our wires is 𝑑.
Any time we’re trying to find the
force per unit length that two parallel current-carrying wires put on each other, we
can use the equation 𝜇 naught 𝐼 one times 𝐼 two divided by two 𝜋𝑑. It doesn’t matter which wire we’re
looking at as we’ll come up with the same equation. If we’re looking for the force per
unit length on wire one, then we’d use a current of 𝐼 one. And the magnetic field would be due
to current 𝐼 two, which once again leaves us with the same relationship.
But what if we have an array of
parallel wires, such as in a solenoid? In this case, let’s say that all of
the wires on the left-hand side have current going into the screen. And all the wires on the right-hand
side have current coming out of the screen. If we plot the individual magnetic
fields around each wire, we can see what happens to the net magnetic field. All the wires on the right-hand
side are gonna have magnetic fields that are directed counterclockwise. The magnetic field we pointed to
the left above the wire will be pointing down to the bottom of the screen at the
left of the wire, will be pointed to the right of the screen underneath the wire,
and be pointing to the top of the screen on the right of the wire.
Between the wires, the magnetic
field’s going to subtract because you have the wire underneath having a magnetic
field to the left of the screen but the wire above having a magnetic field to the
right of the screen, setting up net magnetic field lines around the wires on the
right-hand side that look similar to the ones that we have in yellow going
counterclockwise. Applying the right-hand rule to the
wires in the left-hand side of the screen gives us a field that is clockwise. Above the wire will be directed to
the right of our screen, to the right of our wire will be directed to the bottom of
our screen, below the wire will be directed to the left of our screen, and the left
of our wire will be directed to the top of the screen.
Just as with the previous wires,
the wires on the left-hand side of the screen are gonna have areas where they
interact to make a net magnetic field. In between the wires, these
magnetic fields are going to be pointing in opposite directions, so they’re going to
subtract from each other. This sets up a net magnetic field
as shown in the pattern with the yellow dots, which is essentially a solenoid, where
the magnetic field in between the two sets of wires adds together to make a stronger
net magnetic field. Let’s apply what we’ve learned
about electromagnetic interactions between straight conductors to an example
problem.
The diagram shows concentric
field lines of the magnetic fields of two parallel current-carrying
conductors. Both currents are into the
plane of the diagram. And the currents have the same
magnitude as each other. The increase in radius of the
concentric field lines is constant. And the strength of a magnetic
field at a point around a current is proportional to the perpendicular distance
of the point from the current. Which of the following list of
the points shown in the diagram correctly orders the points from greatest to
smallest magnitude of net magnetic fields? (A) 𝐵, 𝐶, 𝐸, 𝐴, 𝐷. (B) 𝐵, 𝐸, 𝐶, 𝐴, 𝐷. (C) 𝐴, 𝐵, 𝐶, 𝐷, 𝐸. (D) 𝐸, 𝐵, 𝐶, 𝐴, 𝐷. (E) 𝐷, 𝐴, 𝐸, 𝐶, 𝐵.
On the diagram, we see the two
wires set up on the 𝑥-axis along with the five points from 𝐴 to 𝐸. The dotted circles around each
wire represent the magnetic fields. To determine which points are
greater and smaller in magnitude of net magnetic fields, we need to remember two
things. First, when we’re finding the
net magnetic field, we need to use vector addition. This means that if the magnetic
field lines are pointing in the same direction at the same position, we will add
the magnitudes together. And if they’re pointing in the
opposite direction of each other, we will subtract the magnitudes.
Let’s apply this to our diagram
by using the right-hand rule. We need to remember that we can
use the right-hand rule to find the direction of the magnetic field around a
current-carrying wire, where our thumb will point in the direction of the
current and our fingers will curl around the wire to show the direction of the
field lines. When we do the right-hand rule
on our diagram, our thumb points into the screen and our fingers curl
around. This shows that our magnetic
fields will be pointing clockwise around both wires. We can use yellow to show the
direction of the field around the right wire and pink to show the direction of
the field around the left wire.
Let’s look how both fields are
oriented at the five points on our 𝑥-axis. At position 𝐴 and 𝐵, both of
our fields are pointing towards the top of the screen. So the magnitude of the field
will be added together. At position 𝐶 and 𝐷, the
field lines are pointing in opposite directions. Yellow is pointing to the top
of the screen, and pink is pointing to the bottom of the screen. So we would subtract the
magnitudes. Both fields are pointing to the
bottom of the screen at point 𝐸, which means that we will once again add the
magnitudes together. Point 𝐷 is in the middle of
the wire. And since we know that the
magnitude of the magnetic fields is being subtracted at this position, we can
say that the net magnetic field at 𝐷 is zero.
With the net magnetic field at
point 𝐷 being the smallest, we can eliminate any answer choices that don’t have
the last letter as 𝐷. This would be answer choices
(C) and (E). To differentiate between any of
the other points, we need to remember the equation for the magnetic field around
a current-carrying wire. The magnetic field 𝐵 is equal
to the magnetic permeability, 𝜇 naught, times the current in the wire, 𝐼,
divided by two 𝜋𝑟, where 𝑟 is the distance between the position and the
wire. In this example, the current 𝐼
of both wires is the same.
Therefore, no matter where we
are in our diagram, we will have the same current, 𝐼, and our magnetic
permeability, 𝜇 naught, as a constant as well as two 𝜋. Therefore, the magnetic field
will be proportional to one over 𝑟. Looking at our answer choices,
we might be able to solve our problem if the net magnetic field is greater at 𝐸
than at 𝐵 by starting to compare those two points.
Let’s label the wires one and
two to differentiate them, with one being the left wire and two being the right
wire. 𝐸 is one radii, 𝑟, away from
the closest wire, in this case wire two, whereas 𝐵 is just shy of two radii
away from its closest wire, wire one. Comparing the farther wires, 𝐸
is approximately four radii away from its farthest wire, wire one, where 𝐵 is
just shy of five radii away from its farther wire, wire two. Point 𝐸 is closer to the
nearest wire as well as the farthest wire when compared to point 𝐵. And at both 𝐸 and 𝐵, we would
add the magnitude of the two magnetic fields together because they’re pointing
in the same direction.
If we are adding the magnitudes
together and 𝐸 is closer, meaning a smaller 𝑟, there will be a bigger net
magnetic field at 𝐸 than there would be at 𝐵. With the net magnetic field at
point 𝐸 being larger than the net magnetic field at point 𝐵, our final answer
will be choice (D), that the magnitude from greatest to smallest of the net
magnetic field is 𝐸, 𝐵, 𝐶, 𝐴, 𝐷.
Let’s summarize our lesson.
Key Points
Diagrams can be used to
qualitatively determine the net magnetic field around two parallel conducting
wires. To find the force per unit length
that two parallel conductors put on each other, use 𝐹 over 𝐿 is equal to 𝜇 naught
𝐼 one 𝐼 two divided by two 𝜋𝑑. The net magnetic field of an array
of parallel conducting wires can be found by analyzing the individual magnetic
fields.
