Mutual Inductance

Consider two long, thin, cylindrical solenoids, one wound on top of the other. The common length of each solenoid is $l$, and the common radius is $r$. Suppose that the inner solenoid has $N_1$ turns per unit length, and carries a current $I_1$. The magnetic field generated within the inner solenoid is $B_1=\mu_0\,N_1\,I_1$. [See Equation (2.322).] The magnetic flux passing through each turn of the outer solenoid is $\mu_0\, N_1\, I_1\,\pi \,r^2$, and the total flux linking the outer solenoid is therefore ${\mit\Phi}_2 = N_2 \,l\, \mu_0\, N_1\, I_1\,\pi\, r^2$, where $N_2$ is the number of turns per unit length of the outer solenoid. It follows that the mutual inductance of the two solenoids, defined ${\mit\Phi}_2 =M\, I_1$ [see Equation (2.313)] is given by

$$M = \mu_0\, N_1\, N_2 \,\pi\, r^2\, l.$$ (2.342)

Recall that the self inductance of the inner solenoid is

$$L_1 = \mu_0\, N_1^{\,2} \,\pi \,r^2 \,l,$$ (2.343)

and that of the outer solenoid is

$$L_2 = \mu_0 \,N_2^{\,2} \,\pi \,r^2 \,l.$$ (2.344)

[See Equation (2.324).] Hence, the mutual inductance can be written

$$M = \sqrt{L_1\, L_2}.$$ (2.345)

Note that this result depends on the assumption that all of the magnetic flux produced by one solenoid passes through the other solenoid. In reality, some of the flux leaks out, so that the mutual inductance is somewhat less than that given in the previous formula. We can write

$$M= k \sqrt{L_1 \,L_2},$$ (2.346)

where the dimensionless constant $k$ is called the coefficient of coupling, and lies in the range $0\leq k \leq 1$.

Suppose that the inner and outer solenoids have resistances $R_1$ and $R_2$, respectively, If an instantaneous current $I_1$ flow through the inner solenoid then the voltage drop across it due to its resistance is $I_1 \,R_1$. The voltage drop due to the back-emf generated by the self inductance of the solenoid is $L_1\, d I_1/dt$. (See Section 2.3.5.) There is also a back-emf due to inductive coupling with the outer solenoid. The magnetic flux through the inner solenoid due to the instantaneous current $I_2$ flowing through the outer solenoid is

$${\mit\Phi}_1 = M \,I_2.$$ (2.347)

Thus, by Faraday's law and Lenz's law, the back-emf induced in the inner solenoid is

$${\cal E} =-\frac{d{\mit\Phi}_1}{dt}= - M \,\frac{dI_2}{dt}.$$ (2.348)

[See Equation (2.284).] The voltage drop across the inner solenoid due to its mutual inductance with the top coil is minus this expression. Thus, the net voltage drop across the inner solenoid is

$$V_1 = R_1\, I_1 + L_1\, \frac{dI_1}{dt} + M\,\frac{dI_2}{dt}.$$ (2.349)

Likewise, the net voltage drop across the outer solenoid is

$$V_2 = R_2 \,I_2 + L_2 \,\frac{d I_2}{dt} + M\,\frac{d I_1}{dt}.$$ (2.350)

Suppose that, at time $t=0$, we suddenly connect a battery of constant voltage $V_1$ to the inner solenoid. The outer solenoid is assumed to be open-circuited, or connected to a voltmeter of very high internal resistance, so that $I_2=0$. Because $I_2=0$, the circuit equation for the inner solenoid is

$$V_1 = R_1 \,I_1 + L_1\, \frac{d I_1}{dt},$$ (2.351)

where $V_1$ is constant, and $I_1(t=0)=0$. We have already seen the solution to this equation:

$$I_1 = \frac{V_1}{R_1} \left[ 1 - \exp\left(-\frac{R_1\, t}{L_1}\right)\right].$$ (2.352)

[See Equation (2.329).] The circuit equation for the outer solenoid is

$$V_2 = M \,\frac{d I_1}{dt},$$ (2.353)

giving

$$V_2 = V_1\, \frac{M}{L_1} \exp\left(- \frac{R_1\, t}{L_1}\right).$$ (2.354)

It follows from Equation (2.346) that

$$V_2 = V_1 \,k \sqrt{\frac{L_2}{L_1}} \,\exp\left(- \frac{R_1\, t}{L_1}\right).$$ (2.355)

Because $L_1/L_2 =N_1^{\,2}/N_2^{\,2}$ [see Equations (2.343) and (2.344)], we obtain

$$V_2 = V_1\, k \,\frac{N_2}{N_1} \,\exp\left(-\frac{R_1 \,t}{L_1}\right).$$ (2.356)

Now,

$$\frac{V_2(t=0)}{V_1} = k\, \frac{N_2}{N_1},$$ (2.357)

so if $N_2 \gg N_1$ then the voltage in the inner solenoid is considerably amplified in the outer solenoid. This effect is the basis for old-fashioned car ignition systems. A large voltage spike is induced in a secondary circuit (connected to a coil with very many turns) whenever the current in a primary circuit (connected to a coil with not so many turns) is either switched on or off. The primary circuit is connected to the car battery (whose voltage is typically 12 volts). The switching is done by a set of points, which are mechanically opened and closed as the engine turns. The large voltage spike induced in the secondary circuit, as the points are either opened or closed, causes a spark to jump across a gap in this circuit. This spark ignites a petrol/air mixture in one of the engine's cylinders. We might think that the optimum configuration is to have only one turn in the primary circuit, and many turns in the secondary circuit, so that the ratio $N_2/N_1$ is made as large as possible. However, this is not the case. Most of the magnetic flux generated by a single-turn primary coil is likely to miss the secondary coil altogether. This implies that the coefficient of coupling $k$ is small, which reduces the voltage induced in the secondary circuit. Thus, we need a reasonable number of turns in the primary coil in order to localize the induced magnetic flux, so that it links effectively with the secondary coil.