Self Inductance

Consider a long, uniformly wound, cylindrical solenoid of length , and radius , that has turns per unit length, and carries a current . The longitudinal (i.e., directed along the axis of the solenoid) magnetic field within the solenoid is approximately uniform, and is given by

$\displaystyle B =\mu_0\,N\,I.$

(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

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

(2.323)

Hence, the self inductance of the solenoid is

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

(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 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

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

(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.
$\includegraphics[height=2.5in]{Chapter03/fig7_2.eps}$

Suppose that our solenoid has an electrical resistance . Let us connect the ends of the solenoid across the terminals of a battery of constant voltage . The equivalent circuit is shown in Figure 2.27. The inductance and resistance of the solenoid are represented by a perfect inductor, , and a perfect resistor, , connected in series. The voltage drop across the inductor and resistor is equal to the voltage of the battery, . 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, is the current flowing through the solenoid. It follows that

$\displaystyle V = I\,R + L \,\frac{dI}{dt}.$

(2.326)

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

$\displaystyle \frac{dI}{dt}+ \frac{R}{L}\,I= \frac{V}{L}.$

(2.327)

The general solution is

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

(2.328)

The constant is fixed by the initial conditions. Suppose that the battery is connected at time , when . It follows that , so that

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

(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 (which comes from Ohm's law), on the characteristic timescale

$\displaystyle \tau = \frac{L}{R}.$

(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, and **$\tau =L/R$** .
$\includegraphics[height=3in]{Chapter03/fig7_3.eps}$