Question

Two coils, one of radius $0.5\text{ cm}$ having $10$ turns and the other of radius $5\text{ cm}$ having $50$ turns are placed coaxially in air such that their centres are coincident. Calculate the mutual inductance of the two coils.

Hint:

Reciprocity theorem guarantees $M_{12} = M_{21} = M$. The mutual inductance depends only on turns $N_1, N_2$, radii $r_1, r_2$, and medium permeability, independent of operating currents.

Answer 1

Concept:
• Mutual inductance $M$ between two coupled coils relates magnetic flux $\Phi_1$ in coil 1 to current $I_2$ in coil 2: $\Phi_1 = M I_2 \implies M = \frac{\Phi_1}{I_2}$.

• Theoretical formula for concentric coaxial circular coils ($r_1 \ll r_2$) is $M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2}$.

Step 1: Use flux-current ratio
From Part (a), flux linked with smaller coil when current $I_2 = 3\text{ A}$ flows in larger coil is $\Phi_1 = 1.5 \pi^2 \times 10^{-7}\text{ Wb}$.
Mutual inductance $M$:
\[ M = \frac{\Phi_1}{I_2} \]
\[ M = \frac{1.5 \pi^2 \times 10^{-7}\text{ Wb}}{3\text{ A}} \]
\[ M = 0.5 \pi^2 \times 10^{-7}\text{ H} = 5 \pi^2 \times 10^{-8}\text{ H} \]

Step 2: Numerical Evaluation
Using $\pi^2 \approx 9.87$:
\[ M = 0.5 \times 9.87 \times 10^{-7}\text{ H} \approx 4.93 \times 10^{-7}\text{ H} \]

Step 3: Verify using direct formula
\[ M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{2 r_2} \]
\[ M = \frac{(4\pi \times 10^{-7}) \times 10 \times 50 \times \pi \times (5 \times 10^{-3})^2}{2 \times (5 \times 10^{-2})} \]
\[ M = \frac{2000 \pi^2 \times 10^{-7} \times 25 \times 10^{-6}}{10^{-1}} = \frac{50000 \pi^2 \times 10^{-13}}{10^{-1}} = 5 \pi^2 \times 10^{-8}\text{ H} \approx 4.93 \times 10^{-7}\text{ H} \]

Step 4: Conclusion
The mutual inductance between the two coaxial coils is $5 \pi^2 \times 10^{-8}\text{ H} \approx 4.93 \times 10^{-7}\text{ H}$.