Example- Magnetic field of a perfect solenoid
Earlier, we had calculated the magnetic field generated by a long, straight current carrying wire. When we considered an infinitely long wire carrying a current i, that current generated a magnetic field some r distance away from the wire and the magnetic field was equal to Mu zero i over 2 Pi r. We have seen that the magnetic field line geometry was such that they were in the form of concentric circles and then they lie along planes perpendicular to the wire. So at this location, the magnetic field that this current i generated was into the plane. We found, as a result of both [INAUDIBLE] law and as well as Ampere’s law. When we look at this result over here, we see that magnetic field is directly proportional to the current. In other words, if we increase the current, then we end up with a greater magnetic field. Therefore if our purpose is to generate a very strong magnetic field at this point, we simply increase the current flowing through the wire.
Obviously, to be able to do this, in order to increase the current, in other words, through the wire, we apply larger and larger potential difference between the two ends of the wire. The problem with this operation is that we always end up with an upper limit. In other words if we apply a potential difference more than some certain limiting value for a given medium, we can easily break down the electrical properties of that medium. Therefore, by increasing the current, we can increase the magnetic field up to some certain point, but we cannot exceed or go beyond that limiting value.
Then the issue becomes generating a very strong magnetic field by using the same amount of current. Can we do this? And the answer to that question is yes. As a matter of fact, it can be done in a very simple way by taking the wire and putting into a helical form. Something like this. Like a spring. When the current is let flow through this system, as we can see that it will go into the plane at this point, from the cross sectional point of view, and coming out of plane from the bottom part, and it will flow and go into the plane over here and coming out of plane at this point and so on and so forth.
If we look at this picture from the cross sectional point of view, then we’re going to end up with a system something like this. When we cut it down vertically from the cross sectional point of view, we will end up with an upper branch, something like this, and a lower branch. If the upper branch is carrying the current out of plane direction, the lower branch is going to be carrying the current into the plane direction something like this. So in a three dimensional sense, the current is coming out of plane here and going into the plane there. Below the plane, it is going in this direction, coming out, and going into the plane like this and so on and so forth.
Therefore, when we look at the system, we can treat this like the bunch of current loops which are generating a certain magnetic field. If we look at each one of these currents separately, and apply right hand rule, holding the right hand thumb pointing out of plane, we will see that the right hand fingers will circle about the thumb in counterclockwise direction. Therefore on this plane, this current is going to generate magnetic field lines which are circling in counterclockwise direction. So is the next one and then the next one, the next one, and so and so forth.
If you do a similar type of analysis for the ones for the lower branch, which the current is going into the plane, then holding the thumb pointing into the plane and circling the right hand fingers about the thumb, we will see that in this case, the magnetic file lines that this current generates will be in the form of concentric circles and they’re going to be circling in clockwise direction. So is the next one, the next one, the next one, and so on and so forth.
Well, of course these turns are very near to one another and as a result of this, this magnetic field generated by this current over here will overlap with the next one and then that will overlap with the next one, and then that will overlap with the next one. As a result of this, all of the magnetic fields generated by each turn will add to the next one and then they will go back outside of this formation and close upon themselves. As we know that we cannot have open magnetic field lines, they always close upon themselves. Similarly, the lower branch will overlap. The magnetic field generated by each turn will add to the next one and generating field lines something like this. As a result of this formation then, we are going to end up with magnetic field line geometry, that these field lines are eventually closing upon themselves along these directions.
Now we can see that the field line geometry of such a current geometry will be very similar to the case of bar magnets. If we recall, a simple bar magnet with north pole over here and the south pole over here, the magnetic field line geometry of such a magnet was very similar to the field line geometry that we obtain by putting the wire in a helical form of this type. As you recall, for the magnets, we said that the conventional direction of magnetic field lines are such that they emerge from the north pole and enter into the south pole.
In that sense, the formation over here becomes such that this end corresponds to the north pole of the magnet and then this end corresponds to the south pole. As a matter of fact we can easily determine which end will be north, which end will be south again by applying right hand rule. If we just curl out right hand fingers in the direction in which the current is flowing through, which in this case it is going to be in and out perpendicular to the plane of the screen and the current is going to be flowing in counterclockwise direction, and holding the thumb again in open position, we will see that the thumb is going to be pointing to the right. Therefore the right end of the formation will correspond to the north pole. Then the other hand will correspond naturally to the south pole of such a magnetic arrangement.
Well, we call these type of systems as solenoids. Let’s try to determine the magnetic field of a solenoid. Of something like this. By applying Ampere’s law. Magnetic field of a, let’s say, perfect solenoid. What we mean by perfect solenoid is the magnetic field, as you recall, is tangent to the field lines, passing through the point of interest. So inside of the solenoid, we will have the strongest magnetic field because number of field lines passing through per unit area is associated with the strength of the magnetic field and we can easily see that it is the strongest at the ends and as well as inside of the solenoid because that number density will be the same as long as we are inside of the solenoid. Once we leave the solenoid and go away from that, in this region for example, we will see that the tangent b will be something like this. As we go away from the solenoid outside, the strength of the magnetic field will decrease because the number of lines passing through a unit surface will get lesser and lesser.
If we call the length of the solenoid as l and it’s diameter as d, when l is much greater than d, then b out becomes very small that we can treat this like as if it is going to zero. If b out is zero, then we call that type of solenoid as perfect solenoid. Therefore for a perfect solenoid, magnetic field strength outside of the solenoid can be taken as zero. For such a case, let us try to determine the magnetic field inside of this perfect solenoid. In order to do that, we will apply Ampere’s law and as we recall, Ampere’s law was e dot d l integrated over a closed contour is equal to Mu zero times i enclosed.
Okay. Let’s say again l represents the length of the solenoid and capital N represents the number of turns of the solenoid. Small n is the number density of turns and that is simply the total number of turns divided by the total length of the solenoid. To be able to apply this law, we have to choose a closed loop. For this case, let’s say our point of interest is at this location. We’d like to determine the magnetic field at this point p, which is inside of the solenoid and we’re going to choose a closed loop in the form of a rectangle passing through this point p and also it is coinciding with the magnetic field line passing through that point.
Now at this point, the associated magnetic field line is going to be pointing from left to right, something like this. Therefore we choose our loop such that that one side of the rectangular loop coincides with the magnetic field line at the point and here, the magnetic field is going to be tangent to this line. Along this line therefore at this point, magnetic field will be pointing from left to right like this. One side of the loop that we choose coincides with the magnetic field or with the magnetic field line and the other side of the loop outside of the solenoid.
The d l, incremental displacement vector, is along the loop, therefore at this segment, it is going to be like this. Along this segment, it will be like this. An incremental displacement vector for the outside segment is pointing to the left and for this segment is going to be pointing down.
All right. When we look at this loop, we see that it consists of four segments. Namely this is the segment 1, segment 2, segment 3, segment 4. So this is a closed loop integration e dot d l. We can express this as the path integrals along each one of these segments, and finally add them together to get the closed loop integration. In other words, we can express this integral, the left hand side, as integral over the first segment of b dot d l. Let’s write this down in explicit form. For the first segment, b is pointing to the right, d l is also pointing to the right, therefore the angle between them is zero degree, so for this segment we have b d l times cosine of the angle between them and that is cosine of zero. Here I should get rid of this circular sign, because this integral is take over open path or open line over segment 1. In other words, it is not a closed loop integration with this form.
Plus integral of the second segment. When we look at the integral over the second segment, some parts of second segment are inside of the solenoid whereas some part is outside of the solenoid. Therefore we can divide this integral into two segments by saying that the integral from here to there. For that part, we have magnetic field as pointing to the right everywhere inside of the solenoid, so here it is pointing like this for example. The d l is in upward direction, so the angle between them is 90 degrees. Whereas outside, we have zero magnetic field for a perfect solenoid. Therefore, we have b d l cosine of 90 for the part inside of the solenoid so we can call it b inside for that region. Then we also have the segment outside of the second path and for that part, we have b out dot d l, but for this region, b out is zero and as well as inside of the region, since cosine of 90 is zero, there will be no contribution coming from the integral over the second segment.
Moving on, for the integral over the third segment, and this is again outside of the solenoid region, and for a perfect solenoid, magnetic field in that region is zero. Therefore we have b out dotted with incremental displacement vector d l and since b out is zero, there will not be any contribution from the third segment integration also.
Moving on, plus integral over the fourth segment, and the fourth segment is very similar to the second segment. Some part of it is outside of the solenoid and some part is inside of the solenoid. Inside of the solenoid, the angle between magnetic field vector and d l is 90 degrees. Cosine of 90 from the dot product will give us zero. From the outside part of that segment, the magnetic field is zero. Therefore for this region, again, for the inside part, we have d d l cosine 90, and cosine 90 is zero, plus for the same segment outside of the solenoid, since magnetic field is zero, there will not be any contribution again to the magnetic field.
Now if we add these open path integrals over the first, second, third, and fourth segments, then all these path integrals will add up to the closed loop integral over that rectangular loop and that should all add up to Mu zero times i enclosed. If we look at these integrals, only the first one is surviving. Cosine of zero is 1 and as long as we are along this first segment, since it coincides with the magnetic field line, therefore the strength of the magnetic field will be the same everywhere along this line. Therefore the magnetic field magnitude is constant for that part, then we can take it outside of the integral. We are going to end up with b times integral of d l over the first segment is equal to Mu zero times i enclosed.
Now integral of d l over the first segment is going to give us whatever the length of that segment is. In other words, if we add all of these d l’s together, that’s what integration is, along this length, we will eventually end up whatever that length is. Since this is a loop that we choose, we can give it any dimensions that we’d like to and let’s call that length as a for example. In that case, we end up with b times a is equal to Mu zero times i enclosed. So the left hand side of the Ampere’s law is done, now we will try to calculate the right hand side i enclosed.
Now to be able to do that, let’s go back to our diagram. What we mean by i enclosed is the net current flowing through the area surrounded by this loop which is this area. We want to figure out how much current is flowing through this area. The net current flowing through the area surrounded by this closed loop, amperian loop, is what we call i enclosed. Well, of course that current is going to be equal to the amount of turns passing through this surface. If we can express the number of turns passing through this surface, since each turn is carrying the current i, then those number of turns times current i will give us the net current flowing through this surface.
Well, since we know the total number of turns and also total length of the solenoid, then total number of turns divided by total length will give us the number density of turns which we define that as little n. So we will have that much of a number of turns per unit length. Now if we multiply that number density of turns by the length of the region that we’re interested with, which is this region, then that’s going to give us the total number of turns along that length. So little n times a will give us the total number of turns along this length. Since each turn is carrying current i, that number times current i will give us the i enclosed. Therefore here we have is the number density of turns, n, times a, is going to give us the total number of turns along a. That is in explicit form. Let’s say total number of turns divided by total length of solenoid and then times a, this whole thing is going to give us total number of turns along length a.
If we multiply this quantity by i, which is the current flowing through a single turn, then we will get the total number of current passing through the area of interest. This is current through each turn. All right. Now i enclosed is therefore equal to n times a times i. So the left hand side was equal to d a and the right hand we’ll have Mu zero times, for i enclosed we have n times a times i. As you can see, the length of the loop that we choose will cancel on both sides, therefore it is irrelevant how short or how long that we choose the length of this loop. Eventually they will cancel and the magnetic field of a perfect solenoid will be equal to Mu zero times the number density of the solenoid times the current flowing through the solenoid.
We can also express this in terms of the total number of turns of the solenoid and its length. In other words expressing little n as big N over l in this form. Mu zero N over l times i. When we look at this result, it is interesting in the sense that Mu zero is the permeability of free space, that is constant. The total number of turns of the solenoid is a constant quality. Length of the solenoid is a constant quantity and the current flowing through the solenoid is also a constant quantity. So therefore the magnetic field inside of the solenoid is independent of the position and it is a constant quantity. In other words, wherever we go, as long as we are inside of the solenoid, we will see the same magnetic field generated by the current flowing through the solenoid.