Self-inductance

Consider a long solenoid of length $l$, and radius $r$, which has $N$ turns per unit length, and carries a current $I$. The longitudinal (i.e., directed along the axis of the solenoid) magnetic field within the solenoid is approximately uniform, and is given by

$$B= \mu_0 N I.$$ (907)

This result is easily obtained by integrating Ampère's law over a rectangular loop whose long sides run parallel to the axis of the solenoid, one inside the solenoid, and the other outside, and whose short sides run perpendicular to the axis. The magnetic flux though each turn of the loop is $B \pi r^2= \mu_0 N I \pi r^2$. The total flux through the solenoid wire, which has $N l$ turns, is

$${\mit\Phi} = N l \mu_0 N I \pi r^2.$$ (908)

Thus, the self-inductance of the solenoid is

$$L = \frac{{\mit\Phi}}{I} = \mu_0 N^2 \pi r^2 l.$$ (909)

Note that the self-inductance only depends on geometric quantities such as the number of turns per unit length of the solenoid, and the cross-sectional area of the turns.

Suppose that the current $I$ flowing through the solenoid changes. We have to assume that the change is sufficiently slow that we can neglect the displacement current, and retardation effects, in our calculations. This implies that the typical time-scale of the change must be much longer than the time for a light-ray to traverse the circuit. If this is the case, then the above formulae remain valid.

A change in the current implies a change in the magnetic flux linking the solenoid wire, since ${\mit\Phi} = L I$. According to Faraday's law, this change generates an e.m.f. in the wire. By Lenz's law, the e.m.f. is such as to oppose the change in the current--i.e., it is a back e.m.f. We can write

$$V = - \frac{d {\mit\Phi}}{d t} = - L \frac{d I}{dt},$$ (910)

where $V$ is the generated e.m.f.

Figure 51:

Suppose that our solenoid has an electrical resistance $R$. Let us connect the ends of the solenoid across the terminals of a battery of e.m.f. $V$. What is going to happen? The equivalent circuit is shown in Fig. 51. The inductance and resistance of the solenoid are represented by a perfect inductor, $L$, and a perfect resistor, $R$, connected in series. The voltage drop across the inductor and resistor is equal to the e.m.f. of the battery, $V$. The voltage drop across the resistor is simply $I R$, whereas the voltage drop across the inductor (i.e., the back e.m.f.) is $L dI/dt$. Here, $I$ is the current flowing through the solenoid. It follows that

$$V = I R + L \frac{dI}{dt}.$$ (911)

This is a differential equation for the current $I$. We can rearrange it to give

$$\frac{dI}{dt}+ \frac{R}{L} I= \frac{V}{L}.$$ (912)

The general solution is

$$I(t) = \frac{V}{R} + k \exp(-R t/L).$$ (913)

The constant $k$ is fixed by the boundary conditions. Suppose that the battery is connected at time $t=0$, when $I=0$. It follows that $k=-V/R$, so that

$$I(t) = \frac{V}{R} \left[1-\exp(-R t/L) \right].$$ (914)

This curve is sketched in Fig. 52. It can be seen that, after the battery is connected, the current ramps up, and attains its steady-state value $V/R$ (which comes from Ohm's law), on the characteristic time-scale

$$\tau = \frac{L}{R}.$$ (915)

This time-scale is sometimes called the time constant of the circuit, or, somewhat unimaginatively, the L over R time of the circuit.

Figure 52:

We can now appreciate the significance of self-inductance. The back e.m.f. generated in an inductor, as the current tries to change, effectively prevents the current from rising (or falling) much faster than the $L/R$ time. This effect is sometimes advantageous, but often it is a great nuisance. All circuit elements possess some self-inductance, as well as some resistance, and thus have a finite $L/R$ time. This means that when we power up a circuit, the current does not jump up instantaneously to its steady-state value. Instead, the rise is spread out over the L/R time of the circuit. This is a good thing. If the current were to rise instantaneously, then extremely large electric fields would be generated by the sudden jump in the induced magnetic field, leading, inevitably, to breakdown and electric arcing. So, if there were no such thing as self-inductance, then every time you switched an electric circuit on or off there would be a blue flash due to arcing between conductors. Self-inductance can also be a bad thing. Suppose that we possess a fancy power supply, and we wish to use it to send an electric signal down a wire (or transmission line). Of course, the wire or transmission line will possess both resistance and inductance, and will, therefore, have some characteristic $L/R$ time. Suppose that we try to send a square-wave signal down the wire. Since the current in the wire cannot rise or fall faster than the $L/R$ time, the leading and trailing edges of the signal get smoothed out over an $L/R$ time. The typical difference between the signal fed into the wire (upper trace), and that which comes out of the other end (lower trace), is illustrated in Fig. 53. Clearly, there is little point having a fancy power supply unless you also possess a low inductance wire or transmission line, so that the signal from the power supply can be transmitted to some load device without serious distortion.

Figure 53: