Solution: Mutual Inductance of Two Coaxial Solenoids

(a) (i) Definition and SI Unit:

• Definition: Mutual inductance of two coils is defined as the magnetic flux linked with one coil per unit current flowing through the other coil. Alternatively, it is the induced emf in one coil when the rate of change of current in the other coil is unity. [0.5 Mark]
• SI Unit: Henry (H) or $\text{Wb/A}$. [0.5 Mark]

(a) (ii) Derivation:

Step 1 (Governing Law & Setup): [0.5 Mark]

Consider two long coaxial solenoids of length $l$. Let $N_1$ and $N_2$ be the number of turns, and $r_1$ and $r_2$ be the radii ($r_2 > r_1$). When a current $I_2$ flows through the outer solenoid (solenoid 2), the magnetic field $B_2$ produced inside it is uniform:

$$B_2 = \mu_0 n_2 I_2 = \mu_0 \left( \frac{N_2}{l} \right) I_2$$

Step 2 (Mathematical Working): [1.0 Mark]

The magnetic flux $\phi_1$ linked with each turn of the inner solenoid (solenoid 1) due to the field $B_2$ is:

$$\phi_1 = B_2 \times A_1 = \left( \frac{\mu_0 N_2 I_2}{l} \right) (\pi r_1^2)$$

The total flux linkage with the inner solenoid having $N_1$ turns is:

$$N_1 \phi_1 = N_1 \left( \frac{\mu_0 N_2 I_2}{l} \right) (\pi r_1^2)$$

By definition of mutual inductance ($M$):

$$N_1 \phi_1 = M I_2$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Equating the expressions for $N_1 \phi_1$:

$$M I_2 = \frac{\mu_0 N_1 N_2 I_2 \pi r_1^2}{l}$$ $$M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l}$$

The mutual inductance depends on the geometry of the solenoids and the permeability of the core material.

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💡 EL10 Concept Insight:

Think of mutual inductance as a "magnetic handshake": when you change the current in one coil, it creates a changing magnetic field that "reaches out" and pushes electrons in the second coil, creating a current there even though they aren't touching!