GATE 2023 IN Q.12: solution and answer

Official answer: ASource: Official answer key, GATE 2023 IN (IIT Kanpur)

1. Faraday's law: the coil's alternating field at $f=10\text{kHz}$ passes through the copper sheet and induces an emf in it. Copper is a very good conductor ($\sigma \approx 5.8\times 10^7\text{S/m}$), so large eddy currents circulate in the sheet.

2. Lenz's law: these eddy currents set up a flux that opposes the coil's own flux. For the same coil current, less net flux links the coil, so $L=\lambda /I$ goes down.

3. Copper is not magnetic: ${\mu }_r\approx 1$ (it is very slightly diamagnetic), so the sheet gives the flux no easier path and nothing pushes $L$ up. A ferromagnetic sheet such as iron would have the opposite effect.

4. Circuit model (coil = primary, sheet = shorted one-turn secondary): the coil's input impedance is $$Z_{in}=R_1+j\omega L_1+\frac{{\omega }^2{M}^2}{R_2+j\omega L_2}$$ Rationalising the last term gives $$L_{eff}=L_1-\frac{{\omega }^2{M}^2{L}_2}{R_2^2+{\omega }^2{L}_2^2}<L_1,R_{eff}=R_1+\frac{{\omega }^2{M}^2{R}_2}{R_2^2+{\omega }^2{L}_2^2}>R_1$$ So the sheet lowers the effective inductance and raises the effective resistance, because it absorbs eddy-current loss.

5. Check that 10 kHz is high enough: the skin depth is $\delta =\frac{1}{\sqrt{\pi f{\mu }_0\sigma }}$ with ${\mu }_0=4\pi \times 10^{-7}\text{H/m}$, giving $\delta =\frac{1}{\sqrt{\pi (10^4)(4\pi \times 10^{-7})(5.8\times 10^7)}}\approx 0.66\text{mm}$. A typical sheet is thicker than this, so it blocks most of the flux ($\omega L_2\gg R_2$, and $L_{eff}$ approaches $L_1-M^2/L_2$). The drop in $L$ is therefore large enough to see.

6. The LCR meter reports the reactive part of the impedance as an inductance, so it reads less than 10 mH. The reading drops as soon as the sheet is in place and stays there, because nothing makes it recover later. That rules out option (D). Answer: (A).