Self Inductance

Consider a long, uniformly wound, cylindrical solenoid of length $l$, and radius $r$, that 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

$$B =\mu_0\,N\,I.$$ (2.322)

(See Section 2.2.11.) The magnetic flux passing though each turn of the solenoid wire is $B\,\pi \,r^2= \mu_0\, N \,I\,\pi \,r^2$. Thus, the total flux passing through the solenoid wire, which has $N\,l$ turns, is

$${\mit\Phi} = N\,l\, \mu_0 \,N\,I\, \pi \,r^2.$$ (2.323)

Hence, the self inductance of the solenoid is

$$L = \frac{{\mit\Phi}}{I} = \mu_0\, N^2\, \pi\, r^2\, l.$$ (2.324)

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. A change in the current implies a change in the magnetic flux linking the solenoid wire, because ${\mit\Phi} = L \,I$. According to Faraday's law, this change generates an emf in the wire. By Lenz's law, the emf is such as to oppose the change in the current; that is, it is a back-emf. Thus, we can write

$${\cal E} = - \frac{d {\mit\Phi}}{d t} = - L\, \frac{d I}{dt},$$ (2.325)

where ${\cal E}$ is the generated back-emf. [See Equation (2.284).]

Figure 2.27: The equivalent circuit of a solenoid connected to a battery.

Suppose that our solenoid has an electrical resistance $R$. Let us connect the ends of the solenoid across the terminals of a battery of constant voltage $V$. The equivalent circuit is shown in Figure 2.27. 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 voltage of the battery, $V$. The voltage drop across the resistor is simply $I\,R$ (see Section 2.1.11), whereas the voltage drop across the inductor (i.e., minus the back-emf) is $L \,dI/dt$. Here, $I$ is the current flowing through the solenoid. It follows that

$$V = I\,R + L \,\frac{dI}{dt}.$$ (2.326)

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

$$\frac{dI}{dt}+ \frac{R}{L}\,I= \frac{V}{L}.$$ (2.327)

The general solution is

$$I(t) = \frac{V}{R} + k \exp\left(-\frac{R\,t}{L}\right).$$ (2.328)

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

$$I(t) = \frac{V}{R} \left[1-\exp\left(-\frac{R\,t}{L}\right)\right].$$ (2.329)

This curve is shown in Figure 2.28. 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 timescale

$$\tau = \frac{L}{R}.$$ (2.330)

To be more exact, the current has risen to approximately 63% of its final value at time $t=\tau$, and to more than 99% of its final value at time $t=5\,\tau$. The timescale $\tau$ is sometimes called the time constant of the circuit, or (somewhat unimaginatively) the L over R time of the circuit. We conclude that it takes a finite time to establish a steady current flowing through a solenoid.

Figure: 2.28 Typical current rise profile in a circuit of the type shown in Figure 2.27. Here, $I_0 = V/R$ and $\tau =L/R$.