Work against the back emf
While the current in an inductor grows, the back emf pushes against it. The source must keep driving charge through that emf, and that takes work. The work is not lost as heat. It is stored in the magnetic field of the inductor, and the inductor gives it back when the current falls. This is the energy released in the spark at an opening switch.
\[ P = \varepsilon I = LI\dfrac{dI}{dt} \]
Why. the source drives the current \(I\) against the back emf of size \(L\,dI/dt\).
\[ dW = P\,dt = LI\,dI \]
Why. the work done in a short time \(dt\), in which the current grows by \(dI\).
\[ W = \displaystyle\int_0^I LI\,dI = \tfrac12 LI^2 \]
Read it. add the pieces as the current grows from \(0\) to \(I\).
Fig 4.1 shows the sum as an area. The flux linkage \(LI\) grows in proportion to the current, so the graph is a straight line. Each thin strip under it is one piece of work, \(LI\,dI\). The strips fill a triangle, whose area is half the base times the height, \(\tfrac12\times I\times LI\).
Find the energy stored (a) in a \(0.2\,\text{H}\) inductor carrying \(4\,\text{A}\), and (b) in a \(2\,\text{H}\) inductor carrying \(3\,\text{A}\).
(a) \(U = \tfrac12 LI^2 = \tfrac12\times0.2\times16 = 1.6\,\text{J}\).
(b) \(U = \tfrac12\times2\times9 = 9\,\text{J}\).
Values used: the numbers given; no constants are needed.
The formula has the same shape as \(\tfrac12 mv^2\), with \(L\) in place of the mass and \(I\) in place of the speed. It has the same shape as \(\tfrac12 CV^2\) for a capacitor too, from the chapter Capacitance and Dielectrics. A capacitor stores its energy in an electric field, and an inductor in a magnetic one.
A \(0.5\,\text{H}\) inductor carries a steady \(2\,\text{A}\). How much energy does it store, in joules?
Where the energy sits
Put the long solenoid of the last lesson into \(\tfrac12 LI^2\). With \(L = \mu_0 n^2Al\), the energy is \(\tfrac12\mu_0 n^2 I^2\times Al\). The field inside is \(B = \mu_0 nI\), so \(\mu_0 n^2I^2 = B^2/\mu_0\). The energy is therefore \(B^2/2\mu_0\) times \(Al\), the volume of the solenoid. The energy belongs to the field, spread through the space it fills.
The result is general: every magnetic field stores \(B^2/2\mu_0\) joules in each cubic metre. This is the magnetic energy density. A field of \(1\,\text{T}\) holds \(1/(2\times4\pi\times10^{-7}) \approx 4\times10^5\,\text{J/m}^3\). The \(1.5\,\text{T}\) field inside an MRI scanner, like those at AIIMS in New Delhi, holds \(9\times10^5\,\text{J}\) in every cubic metre. Losing that field suddenly would release the energy all at once, so such magnets are run down slowly.
What is the energy density of a magnetic field of \(0.1\,\text{T}\), in joules per cubic metre?
One coil talks to another
Put two coils near each other. A current \(I_1\) in coil 1 sets up a field, and some of it passes through coil 2. The flux linkage of coil 2 is proportional to \(I_1\), and the constant is the mutual inductance \(M\). When \(I_1\) changes, the flux linkage of coil 2 changes, and an emf is induced in coil 2. No wire joins the coils; only the field links them.
In Fig 4.2 the galvanometer on coil 2 kicks only while \(I_1\) is changing. When the switch closes, \(I_1\) grows over about a second. The induced current in coil 2 runs opposite to coil 1's current, opposing the growth of flux. When \(I_1\) is steady, the needle rests at zero, however large \(I_1\) is. When the switch opens, \(I_1\) collapses four times faster. The needle kicks the other way and further, and coil 2's current now runs the same way as coil 1's.
A small coil of \(100\) turns, each of area \(10^{-3}\,\text{m}^2\), sits on the axis inside a long solenoid of \(1000\) turns per metre. Find (a) the flux linkage of the coil per ampere in the solenoid and (b) \(M\). (c) Find the emf in the coil when the solenoid's current changes at \(50\,\text{A/s}\).
(a) The solenoid's field is \(\mu_0 n_1 I_1\). Per ampere, the coil links \(N_2\mu_0 n_1 A = 100\times4\pi\times10^{-7}\times1000\times10^{-3} = 1.26\times10^{-4}\,\text{Wb}\).
(b) \(M\) is that flux linkage per ampere: \(M = \mu_0 n_1 N_2 A = 1.26\times10^{-4}\,\text{H}\).
(c) \(|\varepsilon_2| = M\,dI_1/dt = 1.26\times10^{-4}\times50 = 6.3\times10^{-3}\,\text{V}\), about \(6.3\,\text{mV}\).
A larger \(M\) makes the coils talk loudly. Two coils with \(M = 0.5\,\text{H}\) are one example. If the current in coil 1 falls from \(2\,\text{A}\) to zero in \(0.01\,\text{s}\), the average emf in coil 2 is \(0.5\times2/0.01 = 100\,\text{V}\). The flux linkage of coil 2 changes by \(M\,\Delta I_1 = 0.5\times2 = 1\,\text{Wb}\).
Two coils have a mutual inductance of \(2\,\text{mH}\). The current in coil 1 rises steadily from \(0\) to \(5\,\text{A}\) in \(0.1\,\text{s}\). What emf is induced in coil 2, in volts?
The same both ways
Run the current the other way round: a current in the small coil sends flux through the solenoid. It is not obvious, but the flux linkage per ampere is exactly the same. The mutual inductance of two coils is one number for the pair, whichever coil carries the current. So \(M\) can always be found from the easier direction, usually the coil with the uniform field.
For any two coils, \(M_{12} = M_{21} = M\). The flux linkage of coil 2 per unit current in coil 1 equals the flux linkage of coil 1 per unit current in coil 2.
Mutual inductance cannot be as large as one likes. If all the flux of each coil passes through every turn of the other, \(M = \sqrt{L_1L_2}\). Usually some of the flux misses the other coil, and \(M = k\sqrt{L_1L_2}\). The coupling coefficient \(k\) lies between \(0\) for coils far apart and \(1\) for coils wound tightly together.
Two coaxial solenoids wound one over the other, with the same length \(l\) and cross-section \(A\), show the limit. They have \(L_1 = \mu_0 n_1^2Al\) and \(L_2 = \mu_0 n_2^2Al\). Every line of the inner field threads every turn of both, so \(M = \mu_0 n_1n_2Al\). Its square is \(L_1L_2\), so \(k = 1\).
Two coils have \(L_1 = 0.4\,\text{H}\) and \(L_2 = 0.9\,\text{H}\), with a coupling coefficient of \(0.5\). Find \(M\), and the largest \(M\) they could ever have.
\(M = k\sqrt{L_1L_2} = 0.5\times\sqrt{0.36} = 0.5\times0.6 = 0.3\,\text{H}\).
The largest value is for perfect coupling, \(k = 1\): \(\sqrt{L_1L_2} = 0.6\,\text{H}\).
Two coils have \(L_1 = 2\,\text{mH}\) and \(L_2 = 8\,\text{mH}\), and their mutual inductance is \(2\,\text{mH}\). What is their coupling coefficient?
A small coil sits inside a long solenoid. A current of \(1\,\text{A}\) in the solenoid links \(1.26\times10^{-4}\,\text{Wb}\) with the coil. What flux linkage does \(1\,\text{A}\) in the coil give the solenoid?
A wireless charging pad holds a flat coil with an alternating current in it. The phone has another flat coil inside its back cover. Laid on the pad, the two coils share flux, and the changing current in the pad induces an emf in the phone's coil. A circuit in the phone turns that into a steady current for the battery. Lift the phone a few centimetres and the coupling falls, so charging stops.
- Writing the stored energy as LI.The energy is \(\tfrac12 LI^2\); \(LI\) is the flux linkage.
- Expecting a steady current in coil 1 to keep coil 2's needle deflected.Only a changing \(I_1\) induces an emf in coil 2. A steady current, of any size, gives none.
- Thinking \(M_{12}\) and \(M_{21}\) differ for coils of different sizes.Mutual inductance is one number for the pair of coils.
- Taking k larger than 1.\(M\) can never exceed \(\sqrt{L_1L_2}\), so \(k\) is at most 1.