### Related Formula M = fracphi_2i_1$M = \frac{\phi_2}{i_1}$ B_textstraight wire segment = fracmu_0 i4pi d (sintheta_1 + sintheta_2)$B_{\text{straight wire segment}} = \frac{\mu_0 i}{4\pi d} (\sin\theta_1 + \sin\theta_2)$ ### Core Logic

Assume a current i$i$ flows through the larger square loop of side L$L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side ell$\ell$. The magnetic field at the center of the large square loop (distance d = L/2$d = L/2$ from each side, angles 45^circ$45^\circ$): B = 4 times left[ fracmu_0 i4pi (L/2) (sin 45^circ + sin 45^circ) right]$B = 4 \times \left[ \frac{\mu_0 i}{4\pi (L/2)} (\sin 45^\circ + \sin 45^\circ) \right]$ B = fracmu_0 ipi (L/2) left( frac2sqrt2 right)$B = \frac{\mu_0 i}{\pi (L/2)} \left( \frac{2}{\sqrt{2}} \right)$ B = frac2sqrt2 mu_0 ipi L$B = \frac{2\sqrt{2} \mu_0 i}{\pi L}$ ### Step 2: Mutual Inductance Calculation Flux linkage for the inner loop: phi = B cdot ell^2$\phi = B \cdot \ell^2$ phi = frac2sqrt2 mu_0 ipi L ell^2$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi L} \ell^2$ Given L = ell^2$L = \ell^2$: phi = frac2sqrt2 mu_0 ipi (ell^2) ell^2 = frac2sqrt2 mu_0 ipi$\phi = \frac{2\sqrt{2} \mu_0 i}{\pi (\ell^2)} \ell^2 = \frac{2\sqrt{2} \mu_0 i}{\pi}$ Mutual inductance M$M$: M = fracphii = frac2sqrt2 mu_0pi$M = \frac{\phi}{i} = \frac{2\sqrt{2} \mu_0}{\pi}$ Using mu_0 = 4pi times 10^-7$\mu_0 = 4\pi \times 10^{-7}$: M = frac2sqrt2 times 4pi times 10^-7pi$M = \frac{2\sqrt{2} \times 4\pi \times 10^{-7}}{\pi}$ M = 8sqrt2 times 10^-7mathrm\,H$M = 8\sqrt{2} \times 10^{-7}\mathrm{\,H}$ M = sqrt128 times 10^-7mathrm\,H$M = \sqrt{128} \times 10^{-7}\mathrm{\,H}$ Comparing with sqrtx times 10^-7$\sqrt{x} \times 10^{-7}$, we get x = 128$x = 128$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction