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Example 2- Parallel Wires
Let’s do another example related to the parallel, current carrying wires. In this case, let’s consider a configuration that we have, from the top view, three straight parallel wires. They’re at the corners of a right triangle such that this angle is 90 degrees and these two angles are 45 degrees. Let us assume this one is carrying the current i1, out of plane, whereas this one is carrying the current of i2 into the plane. And let us assume that this one is carrying current i3 also out of plane.
We’d like to figure out direction and magnitude of the net force on i3 generated, or as a result of the currents i1 and i2. Therefore, the question is F-total on i3 due to i1 and i2 is the question mark. To be able to calculate the net force exerted on i3 due to these other current carrying wires, first we have to calculate the net magnetic field generated by i1 and i2 at the location of i3.
In order to do that, first, if we apply the right-hand rule for current i1 keeping the right-hand thumb in the direction of current i1 which is coming out of plane and circling the right-hand fingers about the thumb, we’ll see that the associated magnetic field lines are going to be circling in counterclockwise direction. At the location of i3, the field line will be a circle with this radius, like this, and the the magnetic field will be tangent to that circle. In other words, it’s going to be perpendicular to the radius passing through that point. Therefore, it’s going to be a magnetic field vector pointing in this direction B1, and that’s what it’s going to be along this line of the triangle.
We can give some dimensions to these. Let’s say the length of this segment is R. Actually this will be also R and let’s call this distance as d. By the same token, if we look at the direction of the magnetic field generated by current i2, holding the right-hand thumb now, pointing into the plane direction and circling the right-hand fingers about the thumb, we’ll see that the field lines are going to be rotating in clockwise direction now.
Therefore, at the location of this third wire, the field line passing through that point will be a circle circling in clockwise direction like this at radius R. And the magnetic field will be tangent to that circle, so it’s going to be perpendicular to the radius passing through our point of interest. And that will be therefore, produce a magnetic field in this direction. For simplicity, let’s say the i1 is equal to i2, therefore the strength of these magnetic fields, B1 and B2, will be the same because now i1 is equal to i2. They are same distance away from the point of interest. Therefore B2 is going to be aligned along this direction and it will have the same magnitude with B1.
To be able to get the total magnetic field, we add these two magnetic field vectors vectorially. In order to do that, we introduce a coordinate system such that the point of interest is located at the origin. We resolve these two vectors into their components. Therefore they’re going to have horizontal components like this and the vertical components pointing in upward direction. The horizontal components will be in opposite directions but equal magnitudes. Therefore, they will cancel.
Okay, since i1 and i2 are equal to another, from here we can say B1 is equal to B2 magnitude-wise, and B-total is going to be equal to the sum of the vertical components. So let’s call this direction as x and this direction as y. Therefore B-total is going to be B1y plus B2y, and since B1y is equal to B2y, we can say that B-total is going to be equal to 2 times either B1y or B2y.
To be able express the y component of B1, we need to define an angle. If we define this angle, for example, as angle θ, then B1y becomes equal to B1 times cosine of that angle, cosine of θ, and that will also be equal to B2y. So B-total becomes equal to 2 times B1 times cosine of θ.
Well, again, B1 magnitude is going to be equal to μ0 times i1 over 2π times the distance between the wire and the point of interest, and we call that distance as R. Let’s look at this angle now. When we consider this angle, we see that this side of the angle is — actually we can just look at this angle over here. Since B1 lies along the direction of R along this line, then this angle θ is going to be equal to that angle. And when once we consider this right triangle over here, since this is 90 degrees and this is 45 degrees, θ also has to be equal to 45 degrees because the sum of the interior angles of a triangle always add up to 180 degrees.
Then the total is equal to 2 times μ0 i1 over 2πR times cosine of θ, which is cosine of 45 degrees. Cosine of 45 is root 2 over 2. We can cancel this 2 and that 2. Then we end up with B-total is equal to μ0 i1 over πR times square root of 2 divided by 2.
Once we determine the magnitude of the total magnetic field, then we can calculate the force on this current carrying wire which is carrying the current out of plane direction. We know that F is equal to integral of i dl cross B. So i dl is going to be pointing outside of plane. And from this picture, we see that the total magnetic field is going to be pointing in upward direction.
So i dl cross B, applying the right-hand rule, first keeping the right-hand fingers in the direction of i dl, which is pointing out of plane, and then curling towards the second vector (and that is B pointing in upward direction) i dl cross B is going to generate a force F on this wire pointing to the left. If you try to calculate the magnitude of this force, that’ll be equal to the integral of i dl-magnitude times B-magnitude times sine of the angle between these two vectors.
i dl is coming out of plane and B is on the plane and pointing up. Therefore, the angle between them is 90 degrees. Sine 90 is just 1 and we can take i and B outside of this integral since they’re constant. Once we do that, F is going to be equal to iB times integral of dl integrated over the length of that third wire. And let us assume that the length of all three wires be l. Then this integral is going to give us the length of that third wire, which is equal to l.
Therefore F-total becomes equal to the current flowing through the third wire was i3, therefore we will have i3 over here. i3. And we’re talking about total magnetic field so we have BT over here, i3 BT. And in explicit form, that total magnetic field turned out to be μ0 either i1 or i2 because they are equal to one another, divided by πR times root 2 over 2. And we have the length of the wire times length of the wire 3.
So the force on this third wire due to the other two wires which are carrying the same amount of current in opposite directions, will be equal to root 2 over 2 i times i3 — let’s say i is equal to either i1 or i2 — and we have μ0, permeability of free space, root 2 over 2 times μ0 i times i3 times l and in the denominator we have π and R. So the magnitude of the force is going to be equal to this quantity. If we write this down in vector form, we have introduced this xy coordinate system. As you recall, we call the unit vector along x-axis î and along y as ĵ. Since the force ended up being pointing to the left, in other words a negative extraction, therefore we will multiply this by unit vector –î, indicating that it is going to be pointing to the left or a negative x direction.