 We have a long but thick wire of radius r carrying a current i and our goal in this video is to find the magnetic field Inside the wire now in a previous video We already did this but we calculate for outside the wire and we did that by using Ampere's Circuital law and when we simplified we found the expression for the magnetic field outside to be this Minot i by 2 pi r and for details you can go back and check our previous video But I want to give you a very quick refresher of how we did it We went to some point at a distance r outside the wire and we saw we realized that the magnetic field is in circles And the magnetic field value is the same everywhere along that circle its magnitude and so because we chose an Amperian loop which is also a circle of radius r and because of that over the entire loop B and DL are in the same direction and and B has the same value and so when we simplified B dot DL we just got B DL and Since B is a constant over the entire loop its magnitude is a constant We we pull the B outside the integral and the integral of DL is just adding up all the DLs around along that circle Giving us circumference, which is 2 pi r and according to the law that equals mu naught times the Enclosed current and to find the enclosed current we attached a surface and found that the entire current i is being enclosed And so we get this to be B times 2 pi r equals mu naught into i and if we rearrange we get this expression But again a full detail version can be found in our previous video But now we need to figure out what the magnetic field is inside the wire. So let me get rid of this So this time let's consider a point inside the wire and again, let's say that the distance From here to that point some point inside is r How do I find the magnetic field over here? Well, we can start similar to what we did over here We look at what the magnetic field looks like and just like outside even inside It should be in circles like why would be any different symmetry right so the magnetic field inside will also be in circles So the magnetic field over there Would also be a circle Let's try and draw that Here it goes And according to our right hand thumb rule it's going to be this way And just like before the magnetic field everywhere Is a constant and it's tangential to the circle Which means the amperian loop that we're going to choose just like before Should also be a circle So here it goes Here is our amperian loop This is also a circle We'll also travel in the same direction And if we now calculate what b dot dl in that closed loop is What do you think we'll get now? Can you pause the video and think about this? All right Since b and dl are in the same direction and b is a constant everywhere b dot dl becomes b dl and b can be pulled out integral of dl gives you 2 pi r Exactly the same thing as we got before absolutely no difference. I mean, why would there be difference? The geometry is exactly the same. So the left hand side will directly right now is going to be b times 2 pi r 2 pi Into r No difference whatsoever Hmm, so maybe we get the same answer. Let's see. Let's see What will be the right hand side? Well, right hand side is mu naught times i n closed And again, what is i n closed? You have to attach a surface Dip this imagine dipping it into a soap solution this loop. There will be a film that gets attached Let's attach that here we go And now the current that is penetrating through that surface that becomes your i n closed The question is what is that value? Is that the entire current i? Is that is i the i n closed? Not really Because think about it current i is the total current passing through this entire area Let me show you what I mean if I consider this entire Area of radius r then the total current that is passing through that entire area that is i But our area is less than that right small r is less than that So what we want to calculate i n closed is the total current penetrating that area And the question is how much is that clearly it has to be less than i Right, but how much and how do we figure that out? Hmm Oh one assumption that we can do and let me write that down over here We can assume that current is uniformly distributed over the entire area So let me write that you can assume uniform current distribution So we're assuming that Per square meter or per centimeter square whatever you want the current will remain exactly the same so uniform distribution So given that we're dealing with uniform distribution given you know that in this area the current is i Can you pause the video and think about how much will be the current through this much area? Can you pause and try Okay, now I never used to understand how to do it when there were variables But I felt my brain worked when there were numbers easy numbers So what I would just do is I would just take some random numbers. So let's say The current is I don't know 50 amperes some nice numbers. Let's say this area is I don't know maybe 10 Uh meter square And let's say this area which is a little bit less I don't know maybe it's three meter square and our goal now is To find out how much current is passing over here. I know it's less than 50 amps But how much so how would I do this? Ha now I know see the first thing I'll do is calculate how much amperes I'm getting per meter square So to do that I will just divide the two so 50 Divide by 10 And so I know immediately. Ah, I'm getting five amperes for every meter square for three meter square I just multiply by three and that will give me 15 15 amperes and that's the answer 15 amperes now the answer is not important I just look at this calculation and now I think about how did I do this? So I took the total uh current and divided by the total area that gives me the current per area And then I multiplied it by the area that is needed That is this area So if you didn't if you didn't get this before can you now try and figure out what that current is going to be? All right, so let's do that. So this current is going to be Total current I Divide by this area That area is not r r is the radius. What is the area? It's pi r squared Pi r squared. Let's write that pi r squared So this this gives me the current Amperes per meter square and now if I want in this much area I multiplied by this much area So that is pi small r squared So I get pi small r squared And so notice pi cancels And there we go. Now if I rearrange I'll get my expression. So the magnetic field Inside turns out to be Let me write this mu naught I r squared Oh one r cancels as well Okay, so Yeah, you just get one r divided by 2 pi r squared So now let's digest this formula. What is this formula telling us? Well, it's saying that the magnetic field is proportional to the current more current more magnetic field Very similar over outside as well. So that makes a lot of sense But something interesting that you see now is that the small r is in the numerator meaning as we go further and further away from the center The magnetic field actually increases Notice in the denominator, there is a capital r square and capital r is a constant It's a thickness of the wire that is not change. So as I go further and further away Magnetic field starts increasing. Why is that happening? Because outside the magnetic field starts decreasing. Why why is that? We'll think about it as I go further and further away Clearly the length of the loop starts increasing as you can see over here 2 pi r starts increasing but It'll start the loop will start enclosing more and more current So your current will also start increasing the enclosed current also starts increasing And notice the enclosed current depends on r square. So that increases faster And therefore it all nicely works out That you will finally see that the magnetic field starts increasing as you go farther and farther and farther away So as I go from center center magnetic field will be zero because r is zero So you start with zero as you go farther and farther away magnetic field starts increasing linearly Because b is proportional to r. So if you draw a graph, it'll be a straight line Please draw a graph. Okay. We'll draw a graph. Okay. So it'll be a straight line. Anyways linear starts increasing increasing increasing increasing increasing And then it'll reach a maximum And then if I go outside Now I use this formula. I use this expression And now you will see the magnetic field starts decreasing You may ask why why is it that magnetic field decreases as I go farther away? Because once you come outside now as you go farther away the length of the loop increases But now the total current enclosed stays the same I so this does not increase at all And therefore now as you go farther away the magnetic field drops off as one over r That's the story. And so the last thing I would encourage you to do is Write this entire story in a graph So let's do that. Let me do that somewhere over here. Let me make some space All right. So if you're to draw a graph of magnetic field b Versus the distance r Can you Pause and think about what would that graph look like? All right As long as we are inside the conductor, we will get a linear graph be proportional to r So we'll get a straight line and keep increasing until when until we are inside But how do we represent that in the graph? Well until the r value is equal to capital r Because once r becomes more than capital r. I come outside. So this will be true until r value equals capital r So r equals capital r Now once I go once r value increases more than that now i'm outside And now I have to use this formula and this is b going Decreases as one over r. It's no longer linear. You will get a decrease like this Finally, if you are really really curious you may ask Mahesh, what about when we are at r equals r? On the surface, we know it's a maximum But which formula do I use do I use the formula for inside or do I use the formula for the outside? I would say you can use any one of them. In fact, both of them should give us the same answer And you can check that if I use a capital r The capital r, capital r cancels and I get mu naught i by 2 pi r If I substitute r over here, you get the same answer mu naught i by 2 pi r So on the surface you get mu naught i by 2 pi Capital r and that my dear friends is the story of magnetic field due to long thick conductor carrying currents