9.9 Energy Stored in magnetic field and energy density
In order to calculate the energy stored in the magnetic field of an inductor, let’s recall back the loop equation of an LR circuit. In this circuit, if we consider the rise of current phase, we have a resistor and an inductor connected in series, and once we turn the switch in on position, current i will emerge from the power supply, run through resistor R and through an inductor with an inductance of L from positive terminal towards the negative terminal of the power supply. By choosing a clockwise to traverse the circuit, we have expressed the associated loop equation as ε minus i times R minus L times di over dt is equal to 0.
It was due to the fact that as we cross a resistor in the direction of flow of current, the potential decreases by i times R. And during the rise of current as the current builds up from 0 to i we’re going to end up with a self-induced EMF, and that will show up such that it will oppose its cause. Therefore it will try to generate a current in opposite direction to the direction of flow of this original current. Therefore it’s going to be in a way that we’re crossing an EMF in opposite direction to the direction of EMF arrow as we go through this inductor. Therefore we have –L di over dt, and this was the self-induced EMF part.
Here, let’s go ahead and multiply both sides of this equation by current i. Multiply both sides by current i. If we do that, we will have ε i minus i2 r minus Li di over dt is equal to 0. Let’s rearrange this expression, keep ε times i alone on the left-hand side and move rest of the terms to the right-hand side. Therefore we will have i2 R plus Li di over dt on the right-hand side.
Let’s try to interpret each one of these terms in this equation. As you recall, electromotive force is nothing but a charge pump. It simply pumps the charges with low electrical potential energy to the high electrical potential energy region, and as it does that, it also does a certain amount of work. If it pumping q coulombs of charge through the ε volts of potential difference, then it makes ε times q of work done on q by the seat of EMF. This is, of course, originating directly from the definition of electric potential. Electric potential was the work done per unit charge.
Okay, if we take the derivative of this quantity, then we will have ε times dq over dt, which is going to be equal to ε times i, since dq over dt is i, and that is basically rate of work done on q by ε, but rate of work done is nothing but power. In other words, εi is rate at which seat of electromotive force, EMF, delivers energy to the circuit. So in other words, electromotive force is supplying ε times i of energy in every second to the circuit. That is also equivalent, therefore, power supplied.
Now, the second term over here, therefore εi is the power supplied, and the first term actually on the right-hand side, i2R, is something we are already familiar, and this is rate at which energy appears as thermal energy in the resistor. Rate at which energy appears as thermal energy in the resistor. In other words, that is nothing but power dissipated through the resistor.
Okay, again, if you go back to our equation now, ε times i is the power supplied by the electromotive force to the circuit. In other words, energy supplied to the circuit per unit time. Some of that energy is dissipated per unit time through the resistor. Therefore this much of power is dissipated from that supplied power. Therefore we conclude that rest of the power is going to go the inductor. In other words, this last term on the right-hand side will give us rate at which energy stored in the magnetic field of the inductor.
So we can say then Li di over dt is nothing but equal dUB over dt, which is the rate of magnetic stored in the magnetic field of the inductor, or it is rate at which energy stored in the magnetic field of the inductor. So, dUB over dt is equal to Li di over dt. From here, we can cancel the dt‘s, so dUB will be equal to Li times di. If we integrate both sides, then we will end up with the total energy stored in the magnetic field of an inductor, and that will be equal to — that is constant again. We can take it outside of the integral. And integral of i di is going to give us i2 over 2. So, the magnetic energy of an inductor will be equal to one-half L times inductance times square of the current flowing through that inductor.
So, through inductors again, we can generate magnetic field packages similar to the case of capacitors, which enable us to generate or produce electric field packages. Again, as in that case, we can store energy in the magnetic fields of the inductor, and that energy is going to be equal to one-half inductance of the inductor times the square of the current flowing through the inductor.
Here, let’s make a recall related to the capacitor’s case and say that recall that the energy stored in the electric field of a capacitor was equal to UE, and that was q2 over 2C.
We have defined the concept of energy density earlier, and here also we can define the energy density associated with the magnetic field, the energy density. To do that, let’s consider a solenoid and let’s assume that l represents the length of the solenoid and A represents the cross-sectional area of the solenoid. Therefore A times l is going to represent the volume of the solenoid.
So, we’re considering a solenoid. Let’s say it has a circular cross section something like this, has the length of l and then the cross-sectional area of A, and we have its associated turns, something like this. Well, let’s denote energy density with small uB, and that is by definition total energy of the inductor divided by total volume of the inductor. In our specific case this is going to be equal to UB divided by cross-sectional area of the solenoid times its length, which will give us the volume of that solenoid, a volume through which the magnetic field will fill when certain current i is flowing through the solenoid.
Okay, since the total magnetic energy stored in the magnetic field of an inductor is equal to one-half L, inductance, times the square of the current flowing through the inductor and for a solenoid inductance was equal to μ0n2 times l times A and n2 was the number density of the turns as you recall and, again, l is the length. So, we can express the energy density in explicit form. As for UB, we will have one-half, and the inductance is μ0n2l times A times i2, and divided by the volume, which is A times l.
Here, the length will cancel on the numerator and the denominator, and the cross-sectional area of the solenoid will cancel in the numerator and denominator. We will end up with energy density of a solenoid being equal to one-half μ0n2 times i2. But if you recall that the magnetic field of a solenoid was μ0n times i, and as you recall, this was a constant quantity and it was not changing from point to point inside of the solenoid.
So, in order to have a similar type of expression here, let’s multiply both numerator by μ0 and divide it by μ0. In doing so, we will have one-half, 2 μ0 in the denominator, and multiplying the numerator by mu we will have μ02n2i2, and that quantity is nothing but B2. So, the energy density will therefore be equal to B2 over 2 times permeability of free space, and that expression gives us the magnetic energy density.
And again, you can recall the electrical energy density, which is energy per unit volume for a capacitor, and that was equal to uE is equal to, was equal to one-half ε0 times square of the electric field. Again, we see an interesting parallel between the magnetic field and electric field case.