Inductance
If two coils are placed nearby, the current in one sets up a magnetic field or flux through the other. If the current in the one is changing, then there is a changing magnetic field and flux in the other producing an induced emf and current. In equation-like statements:
Current in one produces flux
in the other.
Changing current in one
produces an emf in the other.
Self Inductance
Further, a changing current in an isolated coil produces an emf in itself. This is called a self-induced emf. The direction of the self-induced emf is so as to produce a current that opposes the current that set it up.
Calculation of the self-induced emf follows from a consideration of the geometry of the coil. Consider a close packed coil with a flux, Φ, passing through each of N turns of the coil. The number of turns times the flux is called the flux linkages, NΦ. The self-induced emf is proportional to the time rate of change of these flux linkages. In the language of calculus, the selfinduced emf is proportional to d/dt of NΦ, so
The quantity NΦ, the flux linkages, is proportional to the current, so make NΦ= Li where L is a constant that depends on how much flux is linked to the coil. Rewriting equation 38-1 with the defined constant L 
The unit of inductance is the Henry which, from equation 38-2, must have units of V.s/A.
The self inductance, or inductance as it is often called, can be calculated for a solenoid. Consider a solenoid of length l having N turns or n turns per unit length. Inside the solenoid the flux is BA, so the number of flux linkages is NΦ=(nl)BA. For a solenoid the magnetic field is μoni, so
The quantity NΦ is equal to the constant L times i so 
The inductance of a solenoid is entirely geometry dependent. Adding a magnetic material to enhance the flux linkages only makes the inductance also dependent on material.
38-1 Calculate the inductance of a solenoid of radius 0.0040m, length 0.015m, and 1.0 × 104 turns per meter.
Solution: 
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The geometric factors n, l, and A (equation 38-3) are difficult to calculate in a closely wound coil, so the measurement of inductance is often done operationally using emf = L(di/dt). Most inductors, or choke coils as they are sometimes called, are not solenoids but short (in length) tightly wound coils. In this geometry it is very difficult to calculate the inductance based on geometry, so this operational method is employed.
38-2 For a coil of 300 turns and inductance 12 × 10-3 H, find the flux through the coil when the current is 5.0 × 10-3 A.
Solution: Use the relationship between flux linkages and inductance NΦ= Li. 
Mutual Inductance
If there are coils within, or adjacent to, other coils, then we can define mutual inductance, the linking of one coil with another via the flux. If the flux is changing, then the current in the one coil is linked to the current in the other coil.
The simplest case to consider is one solenoid-like coil inside another solenoid-like coil. The voltage induced in the second coil is proportional, via the flux linkage, to the changing current in the first coil. The mathematical statement is 
The reverse is also true. The voltage induced in the first coil is proportional to the changing current in the second coil.
The constants are the same. The geometry is the same. These“geometric coupling constants” are again labeled L, the inductance.
38-3 Consider a coil within a coil. The larger coil is 3.0cm in radius, 10cm in length with 1200 turns. The smaller coil is 2.0cm in radius, 4.0cm in length with 50 turns. Calculate the inductance.
Fig. 38-1
Solution: The current in the large solenoid sets up a magnetic field within the coil of B = μ oNi/l. Assume the flux is uniform across the inside of the large coil so that the flux through the small coil is this B times the area of the small coil Φ2 = BA2. The 1 subscript refers to the large coil and 2 refers to the smaller coil. The mutual inductance, in accord with the definition of self inductance is the number of flux linkages divided by the current 
The inductance is totally geometry dependent. For this particular combination of coils
38-4 For the situation of problem 38-3 find the voltage induced in the second coil as a result of a steady change in current of 0.50 A/s.
Solution: 
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38-5 Two coils wound together as a single package have an inductance of 6.0×10-3 H. Find the voltage induced in the second coil when the current is changing in the first at the instantaneous rate of 20 × 10-3 A/s.
Solution: 
Power and Energy Storage
If the emf generated in an inductor is L(di/dt) and the power in an electric circuit is voltage times current, then the general expression for the power in an inductor is
The energy stored in a conductor is the integral over time of the power (remember power is work over time), or in differential form dU = Lidi, so the total energy stored in a coil (the energy is actually stored in the magnetic field) when the current goes from zero to some value i is the integral of this expression 
See Chapter 39, R-L Circuits, for another discussion on this topic.
The energy density in a solenoid is the total energy stored in the solenoid divided by the volume.
Using L = μon2lA,
Or using B = μoni. 
Compare this with the expression for the energy density in an electric field 
38-6 Calculate the energy density in the large coil of problem 38-3 for a current of 0.10 A.
Solution: 