Mutual induction — calculate mutual inductance of two coaxial solenoids

Question

Two coaxial solenoids have the same length $l$ . The inner solenoid has $N_1$ turns and radius $r_1$ . The outer solenoid has $N_2$ turns and radius $r_2$ ( $r_2 > r_1$ ). Derive the expression for mutual inductance $M$ . If $N_1 = 500$ , $N_2 = 1000$ , $r_1 = 2$ cm, $l = 50$ cm, find $M$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

When current $I_2$ flows through the outer solenoid, it creates a uniform magnetic field inside:

B_2 = \mu_0 n_2 I_2 = \frac{\mu_0 N_2 I_2}{l}

This field exists throughout the cross-section of the outer solenoid, including the region where the inner solenoid sits.

The flux through one turn of the inner solenoid (area $A_1 = \pi r_1^2$ ):

\phi = B_2 \times A_1 = \frac{\mu_0 N_2 I_2}{l} \times \pi r_1^2

Total flux linkage with the inner solenoid ( $N_1$ turns):

\Phi = N_1 \phi = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \times I_2

By definition, $\Phi = M I_2$ . So:

M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l}

Notice: $M$ depends on the inner radius $r_1$ (not $r_2$ ), because the flux linking the inner solenoid is limited to its own cross-section.

$\mu_0 = 4\pi \times 10^{-7}$ H/m, $N_1 = 500$ , $N_2 = 1000$ , $r_1 = 0.02$ m, $l = 0.50$ m.

M = \frac{4\pi \times 10^{-7} \times 500 \times 1000 \times \pi \times (0.02)^2}{0.50}

= \frac{4\pi^2 \times 10^{-7} \times 5 \times 10^5 \times 4 \times 10^{-4}}{0.50}

= \frac{4\pi^2 \times 200 \times 10^{-6}}{0.50} = 4\pi^2 \times 400 \times 10^{-6}

= 1600\pi^2 \times 10^{-6} \approx \mathbf{1.58 \times 10^{-2} \text{ H}} \approx 15.8 \text{ mH}

Why This Works

Mutual inductance measures how well one coil’s magnetic field links with another coil. When the inner solenoid sits entirely inside the outer one, ALL of the outer solenoid’s field passes through the inner one (within the inner radius).

The reciprocity of mutual inductance ( $M_{12} = M_{21}$ ) is a deep result. Whether we calculate flux from coil 1 linking coil 2, or from coil 2 linking coil 1, we get the same $M$ . This is known as the Neumann formula.

The coupling coefficient $k = M/\sqrt{L_1 L_2}$ ranges from 0 to 1. For coaxial solenoids with one fully inside the other, $k = r_1/r_2$ .

Alternative Method

You can also compute $M$ by passing current $I_1$ through the inner solenoid and finding the flux through the outer one. The inner solenoid’s field $B_1 = \mu_0 N_1 I_1/l$ exists only within area $\pi r_1^2$ . Flux through the outer solenoid: $N_2 \times B_1 \times \pi r_1^2 = \mu_0 N_1 N_2 \pi r_1^2 I_1/l$ . So $M = \mu_0 N_1 N_2 \pi r_1^2/l$ . Same result — confirming $M_{12} = M_{21}$ .

For JEE numericals, remember the quick form: $M = \mu_0 n_1 n_2 A_1 l = \mu_0 (N_1/l)(N_2/l)(\pi r_1^2) l$ . The key insight is that the relevant area is always the SMALLER cross-section (where the overlap happens).

Common Mistake

Students often use $r_2$ (the outer radius) instead of $r_1$ in the formula. The mutual inductance depends on the area through which the flux is linked, which is the inner solenoid’s cross-section $\pi r_1^2$ . The outer solenoid’s field extends beyond $r_1$ , but the inner coil only “catches” flux within its own area.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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