M = μ₀n₁n₂πr²l for two solenoids. Numerical example.

Mutual Inductance — Two Coaxial Solenoids

Question

Two long coaxial solenoids are wound on the same core. The inner solenoid has $n_1 = 500$ turns/m and radius $r = 4$ cm. The outer solenoid has $n_2 = 300$ turns/m and length $l = 0.5$ m. Find the mutual inductance $M$ of the system.

Solution — Step by Step

For two coaxial solenoids sharing the same core, the mutual inductance is:

M = \mu_0 n_1 n_2 \pi r^2 l

Here $r$ is the radius of the inner solenoid — the one whose flux actually links with the other. The outer solenoid’s radius doesn’t matter because the magnetic field is confined within the inner coil.

$n_1 = 500$ turns/m (inner solenoid)
$n_2 = 300$ turns/m (outer solenoid)
$r = 4$ cm $= 0.04$ m
$l = 0.5$ m
$\mu_0 = 4\pi \times 10^{-7}$ T·m/A

Always convert radius to metres before squaring — this is where most students lose marks.

M = (4\pi \times 10^{-7}) \times 500 \times 300 \times \pi \times (0.04)^2 \times 0.5

Step through this methodically:

= 4\pi \times 10^{-7} \times 1.5 \times 10^5 \times \pi \times 1.6 \times 10^{-3} \times 0.5

= 4\pi^2 \times 10^{-7} \times 1.5 \times 10^5 \times 8 \times 10^{-4}

M = 4\pi^2 \times 1.5 \times 8 \times 10^{-7+5-4}

= 4\pi^2 \times 12 \times 10^{-6}

= 48\pi^2 \times 10^{-6} \text{ H}

Using $\pi^2 \approx 9.87$ :

M \approx 48 \times 9.87 \times 10^{-6} \approx 473.8 \times 10^{-6} \text{ H}

\boxed{M \approx 474 \ \mu\text{H}}

Why This Works

When current flows through one solenoid, it creates a magnetic field $B = \mu_0 n I$ inside the core. This field passes through the cross-sectional area of the inner solenoid, and the total flux links with every turn of the other solenoid.

The key physical insight: mutual inductance tells us how much flux in coil 2 is produced per unit current in coil 1. Mathematically, $M = N_2 \Phi_{21} / I_1$ . Since $N_2 = n_2 l$ and $\Phi_{21} = B \cdot \pi r^2 = \mu_0 n_1 I_1 \pi r^2$ , substituting gives us the formula directly.

Notice that $M$ depends only on geometry — the number of turns per unit length, the radius, and the length. It has nothing to do with the current. This is why we can derive $M$ for a solenoid pair purely from its construction.

Alternative Method — Using Flux Linkage Directly

Instead of using the formula, we can derive $M$ from first principles. This approach is worth practising for JEE because numerical problems sometimes change one parameter and ask how $M$ changes.

Assume current $I$ flows in solenoid 1. The field inside is $B = \mu_0 n_1 I$ .

Total flux linkage with solenoid 2:

\Lambda_2 = N_2 \cdot B \cdot A = (n_2 l) \cdot (\mu_0 n_1 I) \cdot (\pi r^2)

Since $M = \Lambda_2 / I$ :

M = \mu_0 n_1 n_2 \pi r^2 l

Same result, same formula — but now you see why each term is there, not just where to plug numbers.

If the question asks “how does $M$ change if the radius is doubled?” — from the formula, $M \propto r^2$ , so $M$ becomes 4 times the original value. This type of ratio question appears frequently in JEE Main.

Common Mistake

Using the outer solenoid’s radius. Students often use whichever radius is given last, or average the two radii. The formula uses the radius of the inner solenoid only. Why? Because the magnetic flux is confined within the inner solenoid’s cross-section — field lines don’t spread out beyond it. The outer solenoid’s own radius is irrelevant to the flux linkage calculation. In JEE Main 2023, this exact trap appeared — the outer radius was given as 6 cm while inner was 4 cm, and the intended distractor was using 6 cm or some combination of both.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Flux Linkage Directly

Common Mistake

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