Electromagnetic induction > Mutual inductance. Inductance of conductors. Transformers > Problem 11.3.13

Statement

$11.3.13.$
What is the inductance of two long solenoids of radius r1and r2connected as
shown in the figure? The internal solenoid has a length of l1, the external l2.
The number of turns per unit length of the internal solenoid n1, external n2.
Consider cases where the current directions in the turns of both solenoids are
the same and opposite.

Solution

Self‑inductances

For a long solenoid, the total self‑inductance is $L = \mu_0 n^2 \pi r^2 l $
Therefore:

$L_1 = \mu_0 n_1^2 \pi r_1^2 l_1, \qquad
L_2 = \mu_0 n_2^2 \pi r_2^2 l_2$

Mutual inductance

The mutual inductance M is defined from the flux that one coil sends through the other.
The magnetic field created by the outer solenoid (of length$ l_2$) in its interior is uniform and is given by$ B_2 = \mu_0 n_2 I_2$ This field passes through the cross‑section of the inner solenoid ($\pi r_1^2$) along the common length, which is $l_2$
The flux linked by the inner solenoid due to current $I_2 $is:

$\Phi_{12} = (B_2)(\pi r_1^2)(n_1 l_2) = \mu_0 n_1 n_2 \pi r_1^2 l_2 I_2$

By definition, $M = \Phi_{12}/I_2$:

$M = \mu_0 n_1 n_2 \pi r_1^2 l_2$

Total inductance according to the direction of the currents

In a series connection the total inductance is:

$L = L_1 + L_2 \pm 2M$

where the + sign corresponds to currents in the same direction (fields reinforcing each other) and the - sign to currents in opposite directions (fields opposing each other).

Substituting the expressions:

$\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$

Answer

$\boxed{L_{\pm} = \mu_0 \pi \Bigl( n_1^2 r_1^2 l_1 + n_2^2 r_2^2 l_2 \pm 2 n_1 n_2 r_1^2 l_2 \Bigr)}$