Generalization of Amperes Law and Its Application: By Prof. K. Thyagarajan Good morning to all of you. We will continue with our discussion in magnetostatics. You may remember the last lecture we had introduced biot savart law and from biot savart law, we calculated the magnetic field produced by a current loop and the magnetic field produced by an infinitely long straight current carrying conductor. So let me recall. If you have an infinitely long straight current carrying conductor with current (I) passing through the wire, then we calculated the magnetic field at some point P at a distance x from here. So, we call this as x axis, and this is the y axis here and we calculated and showed that the magnetic field B is the magnetic field at this point is pointing inside the paper. So magnetic field is going into the paper here and we are calculated it by using biot savart law. By taking a small current element here and then using that current element calculate the magnetic field at this point and integrate over all current elements. It is interesting to note that all current elements produce magnetic fields in the same direction. So, all we had to do was to add the magnetic fields due to each element and get the total magnetic field. Now we also note that, because of symmetry, the magnetic field will be the same at all points which are at a distance x from here. So, we can generalize this and write that. If I have a current carrying conductor like this and if I calculate the magnetic field at any point which lies on a circle of radius r with the wire at the center, then magnitude of B will be B = the directional magnetic field will be according to right-handed rule. Please note that if the current is going upwards, with the right-handed screw if I move the screw in this direction, then the screw will move up. So, if the current is going upwards, the magnetic field has to curve around the wire like this in this direction. So, this is magnetic field magnitude, independent of distance, independent of z axis along this length of the wire, independent of angle and only depends on the distance of that point from the wire and also note that magnetic field lines form closed lines. Time: -03:19 If this is the current carrying conductor towards me, the magnetic field lines will look like this or closed loops around the current carrying conductor and again, the direction of the magnetic field is determined by the direction of the current that is flowing through this because of the right-handed screw rule, the magnetic fields are in the anti-clockwise direction when the current is coming towards me. So, this also implies that if you take any closed surface suppose I take a closed surface S as many field lines will enter the surface, will go out, and you have this equation, the gauss’s law for magnetic fields

which essentially implies that there are no sources of magnetic field lines that magnetic field lines do not start from any point and at any other point, they form close loops, or they start from here and go to end at infinity. So, this is like the equation satisfied by electrostatic fields, where the flux was equal to . From this calculation of the field produced by a current carrying conductor, we have derived an equation which is the ampere’s law.

So let me recall again this is the current carrying conductor and if I take a circular loop around this point and integrate around this loop. I showed you last time that this is equal to I that is the integral over the loop from across the circular arc. Now this law is always valid. It is very similar to gauss’s law, electrostatics. It is very useful, as I will show you, whenever you can take the magnetic field be outside the integral, then you can actually use this integral formulation to calculate the magnetic field. Otherwise, it is always valid, so what we did essentially was to integrate over the circular path. So let me recall, I have taken small element length dl here. So, if this is angle is dϕ and this radial distance is r, then dl vector magnitude is equal to rdϕ and the magnetic field is also along the same direction as d l vector so is equal to Bdl, which is equal to Brdϕ and magnetic field I just calculated

=

Time: - Now this is a calculation assuming that the wire is at the center of the circular path. Now I want to show you that this value of the integral is always irrespective of the path that I take around the current carrying conductor. So let me again draw a figure here, so this is my current coming out of the plane of the paper, so I take some arbitrary path like this around this current carrying conductor. So let me try to draw a figure here. At this point is perpendicular to this line. This is the line joining the center to this point. is like this and is here. So let me call this angle in between as 𝞱. So let me draw another line here. So, what is .d . . d 𝞱 is the angle supplemented between and d so d is along the path which is not necessarily circular with the wire at the center. is this length and if I call this angle dϕ and this distance r. happens to be equal to rdϕ. So . d

= Time: -09:27 So, if I have a path of integration which is not circular with the wire at the center. What I have shown is that this integral is always equal to (the current in the conductor), which is enclosed by this loop. So, although B and d l are not parallel to each other, happens to be Brdϕ and when I integrate, I just get . How did I choose the direction of integration here? That loop of integration is such that it happens to be along the magnetic field because for a current carrying conductor with current coming up towards me, the magnetic field is anti-clockwise. I can also do an integration in the clockwise direction, for example, if I have a current carrying conductor like this - and if I have a loop like this with integration in the reverse direction, then integral will be minus . Here this is over curve and the same current carrying conductor. If I have another path with like this. Integral is equal to so it depends on the path that you are taking around the current carrying conductor. Time: -10:58 If it happens to be along the direction of the magnetic field satisfying the right-handed rule then in the reverse direction, you can have + or - . If I have not just one conductor but suppose I have more than one conductor carrying current. So let me assume that I have a current carrying conductor with current , another with and I form another some loop like this, so Thus, I can show by choosing many currents that Now, what happens to conductors which are carrying current, but which lie outside the path of integration. So let me take an example here, so I have a current carrying conductor here and I take a path like this. So, what happens now? What I do is, let me draw a line here, so let me say that this corresponds to angle . This corresponds to angle , so Time: - 14:27 So integral along this closed path which does not enclose the current carrying conductor, happens to be zero, so any current element lying outside the loop of integration does not contribute to integral , and that is why I can actually write. So if I have multiple current carrying conductors, so I can write integral is equal to Now couple of things I must mention. I have been drawing curves, which lie in a plane. The path of integration may not lie in a plane, so I can have a wire carrying current like this, so I can integrate some arbitrary path like this, and I will get still get . So, I can have an arbitrary path, some arbitrary path which may not lie a plane, but in the figures which I am drawing here, the curves seem to be lying on a plane. So this is a very general result, so I can draw a figure in which I can say that I can have current carrying conductor another current carrying conductor and another one , so I can have a loop of integration which may be going behind coming like this, so although the curve is not in the plane containing the current, I still have in this integral is equal to Now in this case, as you can see here, this direction corresponds to positive direction with respect to this current, and if I have another current carrying conductor here, for example, does not contribute to this integral or because it is outside the loop of integration. Just like I had mentioned in electrostatics, I must mention here that the magnetic field at every point is determined by all current carrying conductors, just like in the electrostatics case in gauss’s law, the electric field is determined by all charges, while the flux over a closed surface depends only on charges inside. Similarly, the magnetic field produced at any point, for example in this figure, by current , , , , but when I integrate , the only currents that contribute to the integral value are the three currents enclosed by this loop. So please do not forget that magnetic fields at any point are generated by all currents. In ampere’s law, integral , have only the currents which are contained within the loop. Suppose I find a situation integral is equal to 0. This does not imply magnetic field is zero, as we just now saw if the loop of integration exists outside the current carrying conductor integral is zero, although the magnetic field at every point was not zero. Now I want to leave a problem here for you to think. Let me look at it from above, so I have a current coming towards me. 5 amperes, I have another current here going inwards, 5 amperes, another current which is coming towards me, 10 amperes. So let me consider two loops, this one and this one. So, find the value of integral for paths and . Draw paths for which integral is maximum and positive and maximum and negative. So just give some thoughts to this problem and it will help you to understand better the application of ampere’s law. Ok, so I want to apply ampere’s law for certain situations and just like we did for gauss’s law. We obtained gauss’s law and applied gauss’s law to calculate electrostatic fields over charge distributions. Now remember what we found is gauss’s law is always valid. It is useful in certain situations where there is symmetry, because in symmetric situations I can take the electric field out of the integral in gauss’s law, and that will help me to calculate the electric field distribution. Here ampere’s law is always valid. Amperes law is useful whenever I can take the magnetic field outside the integral by some symmetry arguments and use to calculate the magnetic field, so we will start looking at some examples. The first example I want to look at is an infinitely long straight current carrying conductor. Infinitely long straight current carrying conductor: So, this is my current carrying conductor. Now, first thing I notice is, because of symmetry magnetic field cannot depend on z. This distance has to be the same everywhere. It is infinitely long wire. It cannot depend on this angle because if you have a current carrying conductor, this point is the same. At this point I mean it has to be the same. It cannot have an angular dependence. The only dependence it can have r dependence that is the distance from here and magnetic field has r dependence, it has three component, it can have this component, it can have this component and it can have the perpendicular component, a component parallel to the wire. Now I can, for example, think in terms of biot savart law and see that every current element along the wire will produce a magnetic field which is along this direction. No element in this current carrying conductor will ever produce any magnetic field along this direction or along this direction, because please remember the magnetic field direction is so. This is dl vector and this r vector so d l × r in the magnetic field direction. That is always perpendicular to d l and r vector, so it lies like this, so the magnetic field has to be azimuthal. So, if I look from the top in my current carrying conductor, the magnetic field can have only this component. I can also use some symmetry arguments to show that magnetic field cannot have this component, but here I am just using a biot savart law to convince you that the magnetic field, whatever exists, has to be in this direction. Now, once I know the direction of the magnetic field, and once I know that the magnetic field does not depend on this angle, I am going to use this equation. So, this is my ampere’s law. So, what I do is I take a circular path around this wire with a distance r. So, remember at every point, d l is like this, and B is also like this. So, at any point B is parallel to dl vector and so is nothing but Bdl. At this point B is like this, dl is like this at this point. For this integration. I can choose any path, just like gaussian surface. I can choose any gaussian surface. I can choose any curve which I want in this integration. So, my choice is a circular path around the wire at the center, so that it will help me to integrate the left-hand side. If this angle is d ϕ - and this is r dl, is nothing but r dϕ, so so, ampere’s law gives me

B = Please remember, I have obtained ampere’s law by deriving the magnetic field due to infinitely long current carrying conductor. The ampere’s law is a very general law, it’s valid for all situations and I am again using the ampere’s law to calculate the field due to infinitely long current carrying conductor and in the finite long current carrying conductor. I can use some symmetry arguments to find out the direction orientation of B vector and the dependence of B vector on the distance of the point from the position along the wire, the angle with respect to wire etcetera. Time: -26:34 All this, I can do symmetry arguments I can find out, and then I choose an appropriate path of integration, which will help me to take B out of the integral. This is what I have done, so I have taken a circular path around the wire. If I take some arbitrary path, I will not be able to do this, so I must to choose an appropriate path a judiciously chosen path of integration - and here my judiciously chosen path is a circular path around the wire and because I have chosen that path. B happens to be parallel to d l at every point, so I can write So that is a very interesting example which tells me that the magnetic field of an infinitely long current carrying conductor is nothing, but B is equal to , which we had obtained before by using biot savart law. Now you take another example: current distributed uniformly over the cross section of a cylindrical wire of circular cross section and infinitely long, so something like this, I have a thick current carrying conductor. Current is flowing like this in the carrying conductor. So let me assume the radius is r, so the top view will be looking this. I have a circular wire, so current is flowing towards me at every point equally distributed, and so I have to find the magnetic field of this both inside the wire as well as outside the wire. Now I can use the same argument for an infinitely long thin current carrying conductor and say that the magnetic field cannot have a dependence on this position because of infinitely long. So all these points are exactly equivalent, so magnetic field cannot have a dependence on this coordinate, because it’s a cylindrical cross section of a circular wire. Magnetic field cannot have a ϕ dependence and dependence on angle. That means it will be the same everywhere as a function angle. If I take certain distance and calculate the magnetic field at any point along the circle, it has to be the same, because there is no difference between this point. It cannot have a dependence, so magnetic field can only have a r dependence, the distance from the center of the wire. Time: - 30:39 We will all produce a magnetic field which is azimuthal, and which is along this direction, and I can use this immediately to calculate the magnetic field of the current carrying conductor. So, this ampere’s law tells me Inside the conductor (r < R): Now one can show, through symmetry arguments that the magnetic field has to be along the direction of this azimuthal direction, which is tangential to a circular path around the center. So, if I take a path, then . Now I am having a total current I am passing through an area 𝞹 . The total area of the wire is 𝞹 and the current is distributed uniformly across the wire. So, the current I is carried over a wire of cross-sectional area 𝞹 , so I can define what is called as the current density, which is current per unit area, which is I/ 𝞹 . So, if you take a unit area perpendicular to the wire, I will find a current which is passing 𝞹 So the current enclosed by the path is equal to current density into area of which is nothing but I. I/ 𝞹 is current density, multiplied by the area of this enclosed the circular path. so, the current enclosed is . Time: -33:15 So, as I told you, the symmetry arguments tell me that the magnetic field is in this direction along the circular arc so will nothing but Bdl. So, like previous So magnetic field is now proportional to small r. So, the magnetic field at r is equal to zero is zero. The magnetic field increases as you go away from the center. This formula is valid for a path lying inside the conductor, so that’s small part , so that’s for r less than R, because our path is inside the conductor. Outside the conductor (r > R): Now what happens to a path outside the conductor that is r greater than R. I could take a circular path outside, so the current is coming towards me now. The same arguments tell me that the magnetic field has to be along the direction of this circular path, because the circular path has this point at the center. So the magnetic field along the circular path that again tells me integral is equal to integral Brdϕ, which is equal to Br, which is equal to 2𝞹Br. And what is current enclosed is nothing, but I, the total current carried by the conductor, so I get 2𝞹Br = B = This is the same as the magnetic field produced by a current carrying conductor. So, it does not depend on the size of the conductor. Time: - 36:48 So, I have got two expressions here, so let me write it down B = B = Now B (r=R) = So magnetic field is continuous across the boundary. So, if I draw a figure here so this is my current carrying conductor, so this is r. So, look at this formula. Magnetic field at r equal to 0 is 0, so this magnetic field here it increases linearly with r. So, it goes like this up to the point R and then it decreases 1/r. If from 0 to r, magnetic field increases linearly, this is some constant outside and then it decreases is 1/r outside the wire, and so that is the distribution of magnetic field for a current carrying conductor of radius, r and the directional Magnetic field can be obtained by simply looking at the right-hand screw rule and finding out the direction. In this case If the current is coming towards me, the direction of the magnetic field is anticlockwise. Time: -38:35 Coaxial conductor: Now we would like to look at another example: coaxial conductor, so the problem is the following. I have a one conductor here and another one which is outside, so this current flowing like this in conductor and current flowing backwards and so the cross section will look something like this, so current is flow coming towards me, for example, here and current is flowing away from me here. It is uniformly distributed across the cylinder here, so a current I flows from here and flows back from here. It is coaxial conductor because there are two conductors one lying coaxially within the side. This is on the axis of the outer cylindrical conductor. So, what is the magnetic field now? Time: -40:12 Please note that again, because of symmetry magnetic field cannot have a dependence on this position along the conductor. It cannot have a dependence on the angle, so it has to be the same at all points along this angle. It can only have an r dependence and my objective is to find the magnetic field between the points, the radius a and b now. I take a path of integration, so this is my inner conductor. That is outer conductor here, the current is coming towards me here. So here I is away from me and in the central conductor, I is towards me. So, I take a circular path here now, as you can see here, this current is not enclosed by this path, so

The current carried by the inner conductor, or the current carried by outer conductor, but it only has one I and again, because of symmetry you can show that this is 2 𝞹 r B and magnetic field happens to be B = , for a < r < b Now I leave it, you have to find out what is the magnetic field outside the coaxial conductor. Time: -41:21 What is the magnetic field, so I leave with you. Please try to use ampere’s law and to find out what is the magnetic field outside the coaxial conductor pair. It is an interesting problem, and you appreciate this. The coaxial conductors are used in many electronics experiments, and these are very important components of electrical engineering and electronic instrumentation. Now I want to look at another device, which is a very-very important device. Another example is solenoid. Solenoid: Solenoid is a device which has usually a structural cross section and it has current carrying wire bound around it. So let me draw, it is like a coil. These are usually very closely bound coils and I can have a current either flowing upwards or downwards. For example, I can have a current which is flowing downward here in all this. So this wire which comes from here, goes around and finally comes out from here. Current is passing through all the wires, so I take a long wire and wrap it around the cylinder very closely and this is called a solenoid, and this is used to create strong magnetic fields. We define the number of turns per unit length N which means I take a small unit length of this and calculate the number of turns so that is a quantity which I would need to know. If it is closely bound as if each was a circular loop. The structural loops are going like this helix, but if they are very closely bound, I can assume that each winding is a closed loop like this, and these loops are all carrying currents and all loops are carrying the same current. So, my problem is to find out what is the magnetic field produced by this. So, I would like to take an infinitely long solenoid, Infinity long essentially implies that if the radius is a and the length is L and L »a. My length of the solenoid is very large compared to the dimensional solenoid. So, if I have a solenoid and I will be looking somewhere near the center, so for effectively the end effects just disappear. Remember in a capacitor we had the same problem. We had a capacitor with a finite size plate, and we assumed that the plates are infinite extent. Otherwise, I have to look at some end effects and here, so I don’t bother myself with end effects. I have an infinitely long solenoid and I want to find the magnetic field within the solenoid, so I want to use ampere’s law.

To use this law, I must find out what will be depend on coordinates and what will be the direction of B. So, let me draw the solenoid here, this is my solenoid. Now first, I take a surface like this. Now first thing to notice is that because of its infinitely long, the magnetic field cannot have a dependence on this coordinate. It has to be the same every point along this and because it is azimuthally symmetric, I am assuming that is very closely bound coil Time: -44:28 It cannot have a ϕ dependence, it can only have a dependence on r. It cannot have a dependence on z, it cannot have dependence on phi. So let me take a surface like this, so this is the top surface. This is the closed surface and I know that magnetic field satisfy this equation. 0 This is my lower surface here and upper surface is here. So let me call this , this is .. Now please note that, because the normal to this surface is like this, the d on this surface is pointing up and, on this surface, pointing down. When you do an integration like this, the d is an area vector with outward normal, so d here upward, d vector here is downwards: magnetic field is independent of this distance, so you can immediately understand that the flux that is exiting from the upper surface must be exactly equal to the flux that is entering from the lower surface, because remember the normal are oppositely oriented. So if the magnetic field happens to be pointing upwards, then as much flux is entering here as is getting out here because magnetic field here and here are the same. Areas are the same, so they have to cancel each other, so the integral area over and just cancel off whether the magnetic field is pointing downward or upward or any angle, because there is no dependence on this position as much flux is entering or leaving the lower surface, as much as leaving or entering the upper surface. So the only integral which will be left will be over and because there is no dependence of magnetic field along the direction. Let me look at the top surface here, so this is my solenoid, and I am taking a path like this. This is my path on the surface, so this is the direction. A normal here is like this. So, what I will get is

= 0

There can be no radial component of the magnetic field. Magnetic field cannot have a component pointing away from the solenoid, so I have used gauss’s law for magnetic fields to show that magnetic field cannot have a radial component for the solenoid. It is closely bound infinitely long solenoid. Please remember I am calculating the magnetic field for an infinitely long closely bound solenoid. Time: -49:28 Now let me do take another integration, so this is my solenoid and I take a circular path like this, and I want to use this equation: ampere’s law. If I look from the top that is my solenoid and I am taking a circular path like this. Now that is a solenoid and my path is like this, and this is r. Now so because my path is like this. If I call this as a component , this is dϕ. is the ϕ-component of magnetic field, which is the azimuthal component, which is along the tangent to the circle. Now please note that in my solenoid I am assuming very tightly bound solenoid, so the coil is like this. My curve goes like this and if you look at this curve, there is no current entering or leaving this path, because the current is lying inside here, and the other currents are not crossing the path. There is net current, which is entering or leaving this circular path, which is perpendicular to solenoid. Solenoid is like this, and my path is like this. So, the right-hand side, current enclosed will be zero and because the integration is along this curve, I will get 2 𝞹 r is the circumference of the circle. So This implies there can be no azimuthal component. If the solenoid is like this, first thing I showed you is there can be no component of magnetic field like this. Time: -52:22 I have also shown you there can be no magnetic field component in this direction, so you can know magnetic field component like this. In fact, you can use biot savart law with current elements in different directions to find out again using symmetries, to tell you finally that magnetic field has to have only component , so z axis is like this. For the solenoid, it cannot have a component like this, it cannot have a component like this which I showed you first here. I showed you that my solenoid is like this, so it cannot have an r component. It cannot have a component away from this solenoid. It cannot have a component along the azimuthal direction. The only component left is this exact component ( . That is the only component which can survive. Now having obtained this, I can use ampere’s law to find out the magnetic field of a solenoid. So let me draw the solenoid again here, so this is a section of the solenoid. So, the current is coming towards me, and this is entering here. These are the coils. Now first thing I do, take a loop outside. So please remember . B can only have z- component and it can only depend on the radius r. So now I use this ampere’s law for this loop.

Now this loop does not enclose any current, so this must be equal to zero. So,

This integral must be zero and this term of zero. Because remember, the path of integration is like this, and the magnetic field can only have a z component, so these two integrals (b to c and d to a) are zero and magnetic field does not depend on the position at all. So, if I call this , this is . Then B ( B ( So magnetic field seems to be independent of the distance from the axis. Another result we have got so what I will do is I will stop my lecture here. In the next class, I will continue with this discussion, and we will calculate with all these arguments. What is the magnetic field inside the solenoid and outside the solenoid? Time: -55:43 Let me see here that I know that the magnetic field at infinite distances from the solenoid must be become zero. So, if I take to infinity, it must be becoming zero, so B outside the solenoid is zero, that is something we have obtained today. In the next class I will take another amperian loop and I will calculate and show you that the magnetic field within the solenoid is uniform and will calculate the magnitude of the magnetic field. Thank you.