8.7 Current Carrying Parallel Wires
All right. Earlier we have seen that a moving charge or bunch of moving charges, which is electric current, is the source of magnetic field. In other words, a single moving charge or an electric current generates the magnetic field. If we place any other moving charge qv or a current carrying conductor inside of this region, then this external magnetic field exerts a magnetic force on that moving charge or moving charges.
First we studied the first part of these interactions and looked at how the magnetic field is generated from its source and first we have introduced and experimental law which was called Biot-Savart law and it was given for a single charge that the strength, that the magnetic field, generated by a single moving charge is equal to permeability from free space divide by 4π times qv cross unit vector r̂ in radial direction divided by the square of the distance at that instant between the moving charge and the point of interest.
Or for a bunch of moving charges, in other words, a current carrying wire, the magnetic field was equal to μ0 over 4π. Integral of i dl cross r over r3. We ended up with the r3 term once we express this unit vector in radial direction in terms of the position vector r.
Then we went ahead and introduced another law, which was called “Ampere’s law” in order to calculate the magnetic field. In this case the law was given as B, magnetic field, dotted with incremental displacement vector dl integrated over a closed loop, c, is equal to permeable free space near 0 times the net current flowing through the area surrounded by this loop.
Now, these two laws helped us to calculate how the magnetic field is generated from its source and the second part of this, or these interactions, the force exerted by the external magnetic field on a moving charge, which we called it “magnetic force”, calculated from qv cross B. So the the moving charge qv, the vector cross with the external magnetic field vector gave us the force generated by that external magnetic field. And of course to be able to have this force, the charge had to be moving. If it is at rest, in other words, if v is 0, naturally the magnetic field will not generate this magnetic force on that charge.
And furthermore or so while looking at the properties of this force, we said that if the charge is moving, either parallel to the magnetic field or antiparallel to the magnetic field, for those cases or so, magnetic field will not generate a magnetic force on that charge. Because remember we looked at magnitude of this force, which is qv magnitude times sine of the angle big B. And since sine 0 or sine 180° is to equal 0, then the force acting on that particle will be equal to 0.
For the bunch of moving charges moving along a wire for an electric current, then this force was equal to integral of i dl cross B. In other words, we looked at the force exerted on a very small segment of the wire due to this external magnetic field, and then we looked at the next segment and the next segment and obtain all those forces and then we add them vectorially and the addition process is nothing but integration along the length of that wire. Then we ended up with the total force.
Now we are going to look at, consider an example, which will be dealing with the both parts of these interactions. Let’s consider two parallel, straight current carrying wires. In this case, we have two long, straight wires like these, parallel wires. And let’s assume that they are carrying the currents ia for the first one and ib for the other one. In upward, both of them are, in upward direction, and also let’s assume that length of these wires equal to l and they’re separated from one another by a distance of d.
So here we know that each one of these wires will generate their own magnetic field and the magnetic field generated by current a is going to be an external magnetic field for current b. In other words, then the magnetic field of current a is going to exert a force on current b. Similarly, the current b is going to generate its own magnetic field and that magnetic field will be an external magnetic field for the current a. So then that external magnetic field associated with current b will exert a force on current a.
If we look at the direction of these forces, for current a, holding the right hand thumb in the direction of flow of current and curling the right had fingers above the thumb, we will see that that is going to generate concentric magnetic field lines and since the current is going in upward direction and those field lines are going to be rotating in counterclockwise direction. Therefore, in three-dimensional sense, as they go about this wire in counterclockwise direction, they are going to be coming out of plane out the left hand side of this wire and going all way and going into the plane at the location of the current b or wire b. Therefore the magnetic field generated by current a is going to be pointing into the plane at the location of current b.
Now, since this is an external magnetic field, relative to the current b, it will generate a force, qv cross B force, or I shall say, i dl cross B force on this wire. And i dl is an element in the direction of flow of current. Therefore it is going to be pointing in upward direction. B is into the plane, into the plane, therefore i dl cross B is going to generate a magnetic force, pointing to the left. And this is going to to be the force on wire b due to wire a. This is our wire b and this is our wire a.
In a similar way, if we continue and apply the same procedure for current b, current b is going to generate its own magnetic field. Again, holding right hand thumb in the direction of flow of current, the associated magnetic field lines are going to be in the form of concentric circles rotating in counterclockwise direction. So these lines are going to to be going into the plane on the right-hand side and coming out of the plane on the left-hand side. At the location of wire a, the magnetic field generated by the current b is going to be coming out of plane. And this is the magnetic field Bb.
This magnetic field will be an external magnetic field for this wire and therefore it’s going to generate i dl cross B force on every segment of wire a and i dl is, again, in the direction of flow of current, up. B is coming out of plane. i dl cross B is going to generate an incremental force of dFab and all this force will be in the same direction for all these segments. Therefore when we add them the net force is going to be pointing to the right and that is the force on a due to the current b. And obviously directed from action reaction principle, we will expect that these two forces to be in equal magnitude. If we look at the nature of these forces we can easily see that they are going to be in an attractive manner.
If we consider the case that the current flows in opposite directions along these two parallel wires, then applying the same procedure or same analysis, let’s say, if the current a is going upward direction while the current b is flowing through wire b in downward direction. Current a, again using right had rule, will generate a magnetic field Ba which is going into the plane at the location wire b. And current b will generate a magnetic field, holding right-hand thumb pointing in downward direction and circling the right hand fingers in the direction of about the thumb in about the wire we will see that the magnetic field lines are going to be coming out of plane on the right-hand side and going in to the plane on the left-hand side. Therefore at the location of the wire a, magnetic field generated by current b is going to be into the plane direction.
Now if we look at the direction of the force generated by these external magnetic fields on these wires. For the force exerted on current b due to the wire, or current a, i dl cross B. i dl is going to be in downward direction. B is into the plane and applying right hand rule, i dl cross B will generate a force to the right and this is going to be the force on b due to a. Similarly, if we consider the current a, which is inside of this external magnetic field B, i dl cross B is going to generate a force to the left, like this, and this is a force on a due to b.
So in this case, the nature of two forces is basically repulsive. Therefore, here we can say attractive forces, once these two wires are carrying the current in the same direction and if the current is flowing in opposite directions along these two parallel wires, then we end up with repulsive nature of forces. One can actually observe the existence of these forces if we suspend two wires and let the current flow in the same direction on both of them, we will see that they’re going to move towards each other under the influence of these attractive forces. On the other hand, if we let the current flow in opposite directions then we will see that they will move away for one another.
All right, now from Newton’s third law, we know that the magnitude of these forces should be equal to one another and lets call this magnitude as F. Since F is equal to integral of i dl cross B, that’s the magnetic force, and if we just calculate either Fba or Fab magnitude for any one of these cases, that will be equal to integral of i dl. So if we consider this wire over here, i dl is going to be in upward direction and B is into the plane therefore the angle between them will be 90°. And since is i is constant and B is constant we can take it outside of the integral. Sine 90 is just 1.
Then we end up, F is just equal to iB times integral of dl integrated over the length of this whole wire and we call that length as l. Therefore this integration is going to give us just l, so the magnitude of this force is going to be i times B times l.
If we express the magnetic field, and of course when we consider the force Fba — so lets call this on Fba, then the current flowing that wire is ib and the magnetic, external magnetic field is generated by current a and this is therefore Ba. Then if we carry on those subscripts, we will have ib, Ba, and here we will have ib, and Ba times l. If we write down the explicit form of Ba, the magnetic field generated by this straight, current carrying conductor, and we have calculated for such a long, straight current carrying wire, earlier by applying either Biot-Savart law or Ampere’s law, we have found out that this is equal to, that magnetic field is equal to μ0 times the current, which is ia, divided by 2π times the distance from wire to the point of interest. So from this wire to the point of interest, or the distance between the point of interest and the wire is simply equal to d. Therefore we will have 2πd over here.
So we force on wire b due to wire a is going to be equal to ib times Ba and that is μ0 ia over 2πd and times l. Of course this is also equal to force, magnitude-wise, on a due to wire b. The magnitude of these forces for both of these two cases will be the same and their directions are going to be, depending upon how the current is flowing through these wires, whether it is in the same direction or in opposite direction, then we will end up with either attractive or repulsive forces.