 on self inductance of solenoid. Solenoid is like the most basic one, the most basic practical practically used device when it comes to magnetism. So, we will talk about solenoids self inductance. In order to have capacitance does it need to have a charge? An empty bucket does it need to have water to have any capacity? Similarly, capacitor or inductor, inductor do not need to have current for inductance, capacitance do not need to have charge for capacitance but when you try to find out the capacity of a bucket you pour water in it. Similarly, when you try to find out capacity of capacitance you pour charge in it and likewise when you find out the capacity or sorry inductance you have to assume a current in it, it need not be there. So, assume that you have a solenoid coil like this. The length of the solenoid is l, the radius of the solenoid is r, number of turns per unit length of a solenoid is small n. Try to find out self inductance of this coil. The magnetic field inside the solenoid will be how much? Mu naught n, if current is there then n i, no current no magnetic field. Assume there is a current i. So, magnetic field will be mu naught n i, which direction the magnetic field will be? Magnetic field will be uniform along the axis like this. So, how much will be the flux because of this? What, what, what? i is equal to n b a. So, d b mu naught n i into the sense of magnetic field. So, how much that is mu naught n i into? Area of the whole thing. How much? You are confused. You have to take area, which is perpendicular to the magnetic field like this, which crosses there. So, area is pi r square and number of turns are n is number of turns per unit length. Capital n is not given. So, small n into l n, fine. So, this will be what? Mu naught n pi r square or mu naught n square pi r square l times i, fine. So, this bracket term is self-inductance or not? The way self-inductance is defined, the bracket term is self-inductance. Look at it. It depends on geometry, number of turns, radius, length. Just changing the geometry, you can change the inductance. And one more thing you should remember, one thing you should remember that I can put a soft iron core inside, like I can change the material, which is inside. So, just like you have learned about dielectric in electric field, you can change the medium and change magnetic field also changes. Just like electric field becomes e by k and epsilon naught becomes k times epsilon naught. If you change the material over here, mu naught becomes mu r into mu naught. And some of the material have magnetic susceptibility very, very high, like mu r may be thousand. So, magnetic field can become thousand times also, fine. So, you can change the material. So, mu naught changes. If you change the material, mu naught becomes mu r into mu naught. Mu r is relative permeability of a medium. That we are going to study in magnesium matter chapter, but I am just telling you so that you should keep in mind. So, this is about self-inductance. Let us see the mutual inductance. But this is see, this is just one example of self-inductance, alright. So, somebody ask you what is a self-inductance, how much it is? You should ask about what? For what? Fine. So, we have derived self-inductance of a solenoid only, nothing else. Write down mutual inductance. So, we just take one example of mutual inductance. We have a solenoid like this. We need another coil, right, because we are talking about the mutual inductance. So, let us say this is R1, number of turns are N1, the length here is L only and inside this coil, there is another one. Its radius is R2, number of turns per unit length is N2. Fine. Now, try to find out the mutual inductance for this arrangement. Current in one coil creates flux in the other coil. The constant proportionality is the mutual inductance, how much it is? Assume the current in the first coil or like here only, let us say this is I1. How much is the flux in the second coil? The L is same. Current is in the coil one, you have to find flux in two. Magnificent will be how much? B will be how much? Because of I1. Do you know which end? N1, I1. This is B. So, flux in the second coil is how much? This into which area? Which is pi R2 square. Then number of turns of? Second coil. Second coil which is? N2 L2. So, you will get U0 N1 N2 pi R2 square L2 into I1. This is pi U. So, this bracket term is the mutual inductance. Getting it? How we found out this? By assuming current in the first coil and the flux in the second coil. Can you do it in a reverse manner? Assume current in the second coil and find out flux in the first coil. Will you get the same mutual inductance? You should get R1 square. I2 is fine because I2 can be taken out. You should get R2 square actually. See first of all what is the magnetic field? Mu0 N2 into I2. So, when you find the flux, you have to find the flux in the bigger coil. Is there a magnetic field over here? Outside the first coil? Outside the second coil for you? There will be magnetic field over here or not? No. Outside the sauna there is no magnetic field. So, even though you are finding the flux on the bigger this thing, but magnetic field is only in this area. So, you have to multiply only that area in which the magnetic field is. So, this is Mu0 N2 I2 pi R2 square into N2 into not N2, N1. You have to find the flux in the bigger one. N1 into N2. So, you get the same thing. Mu0 N1 N2 pi R2 square L into I2. This is the mutual inductance. Both ways you get the same thing. Went on a question. Two concentric sulfur coils. One of radius R1, other is radius R2. R2 is very less compared to R1. R2 is very less compared to R1. They are placed co-actually with their centers coinciding. Find out the mutual inductance for the surrender. Find out mutual inductance. Mutual is mutual. I2 is so less it can be considered as a point. Very less. You know that magnetic field at the center is how much Mu0 I divided by 2 times R for a circular coil of radius R and having current I use this. Assume current in the bigger coil and find the flux in the smaller coil. This is R1 and this radius is R2. There you go. Magnetic field at this. Assume there is a current on the bigger one. You have to assume a current. So, magnetic field will be this much which is how much? Mu0 I1 by 2 R1. This is the magnetic field and the flux on the second coil will be how much? Magnetic field into area of the second coil. The area is so less you can assume it to be constant. Magnetic field to be constant throughout the area. So, you can say that the flux is Mu0 I1 2 R1 into pi R2 square. So, everything apart from the magnet becomes the self magnet. This magnetic field is at the center. If you go away from the center this should change, but the coil is so less in size you assume it to be constant throughout the area. Fine. So, little bit more is left about the inductance. See, usually we ignore mutual inductance. We just take self inductance. So, assume you have a coil like this. You know that flux is equal to L into current in the coil. What is the current in the coil? Whatever is going here, the current might be going from a battery. Getting it? This is flux. So, emf is what? Minus of d phi by dt which is equal to minus of L di by dt. Now, where this emf is in this circuit? Where it is? Where do you see this emf being created in the circuit? Across the coil. It is the emf on the coil itself, but if you have a DC source here, after sometime current becomes constant like immediately actually. Current becomes 0. So, di by dt will become 0. So, emf will become 0 for the coil. But if the switch was open earlier and you close it suddenly current rises. So, there is di by dt which is huge. So, if di by dt is huge, even the emf produced in the coil is huge and emf will be produced in such a way. See this negative sign. It will be created in such a way that it will oppose this emf or it will oppose the increase in the current. Getting it? So, a back emf gets created if the current tries to increase. The faster the current tries to increase, the stronger will be the back emf. And what this back emf will do? It will not let the current increase very fast. So, it acts like a spring. A spring is a damper. It is there in the bike, a shock absorber anywhere. It does not let force to be felt suddenly. It does not let impulse to be felt. Similarly, the inductor will not let current increase very quickly because it will generate back emf. So, it is like a spring of electricity. So, many times we will have inductor and resistor circuit where if you apply Kirchhoff's law, this you have to take as emf. For example, suppose see we are going beyond syllabus now Suppose you have an inductor L and you connect a resistance across it like this. Find out current as a function of time. Apply Kirchhoff's law. Assume current I is flowing. So, if you assume current I to be flowing, this is potential difference V which will apply constant potential difference. So, V minus L di by dt minus I times R is equal to 0. This is Kirchhoff's equation. Potential difference across the inductor minus potential difference across the resistance. It is a loop equation, loop rule. So, you will get V minus I R is equal to L di by dt. So, L di divided by V minus I R is equal to. So, when you integrate this from 0 to T, current goes from 0 to I. It becomes minus L by R log of V minus I R 0 to I is equal to T. So, you will have ln of V minus I R minus ln V is equal to minus RT by L. So, you will have ln V minus I R divided by V is equal to minus RT by L. So, V minus I R by T is e to the power minus RT by L. So, you will get current is equal to V by R 1 minus e to the power minus RT by L. So, this is how the current will grow in this circuit. Let me know if you have any doubt. Just read it once. Every time you see inductor potential difference is this and you can use Kirchhoff's loop rule to write an equation like this. No doubt. Look at here. When T is equal to 0, what is the current? 0, 0. It becomes e to the power 0. 1 minus e to the power 0 is 1. So, current is 0. Fine. When T is infinity, what will happen? This term will become 0 e to the power minus infinity. So, current will become V by R. So, after some time, when T becomes large enough, you will gain steady current and the current will grow like this. It will grow like that and it will reach a maximum ideal. This is an asymptote V by R. See how gradually it rises. If inductor would not have been there, it will suddenly jump and that could be harmful for the equipments. This is current and time. V by R. Maximum current. Now, we talked about inductor being a device that stores the magnetic energy but we have not yet found out how much is that energy. So, now let us try to find out how much is that energy. In capacitance, we have defined capacitance and also quantified how much energy capacitor will have. Similarly, let us try to derive for inductor. So, inductor will have how much emf minus l d i by t t. So, suppose the current is growing, it will have this back emf. Now, in order to find out the energy inside the inductor, what we have to do? We have to find out how much work you are doing to transport the charges from here till here. Whatever work you have to do, that work is getting stored as energy. It is like simply like in capacitance also, we have found out charge times potential difference. This is the small work you have to do. Similarly, here also, small amount of work you have to do is minus of d q into l d i by d t. This minus is symbolic. So, you can just ignore it because you have to do, you have to push the charge this way but you are getting pushed in opposite direction. So, the solenoid is doing negative work but you are doing positive work. So, w is what? l into d q by d t into d i. What is d q by d t? So, d w will be equal to l into i d i. I have to reduce it into one variable otherwise I would not be able to integrate. So, total work done will be what? You have to integrate now from 0 to i work done 0 to w. So, work done will be equal to half l i square. This is also the magnetic energy stored in an inductor. So, remember we found magnetic sorry electrostatic energy capacitance as half c v square. Then we have also found out electrostatic energy in form of electric field half c v not e square was energy density you remember? No, if you do not remember do one thing try to find magnetic potential energy in the form of magnetic field as in how much this will be for solenoid for a solenoid of number of turns n length l radius r. How much is this? You have to find in terms of b magnetic field. Potentially as a function of b only b n of l n r. Current should not be there i should not be present. So, you know b is equal to mu naught n i. So, i is equal to b divided by mu naught n. So, you can substitute for current. What is l? So, energy is half l i square which will be half this n square get cancelled 1 mu get cancelled you will get b square by 2 mu naught multiplied by pi r square l. What is pi r square l? The volume right. So, potential per unit volume is b square by 2 mu naught. So, we have found out magnetic potential energy per unit volume or you can say magnetic potential energy density in the form of magnetic field b. The electrostatic potential energy density u e was half epsilon naught e square this is b square by 2 mu naught fine. So, this is what this chapter is all about any doubt from this chapter. You have not practiced anything. So, you have to finish