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(A) According to Faraday's Law,the induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\Phi_B}{dt} = -A \frac{dB}{dt}.$ Since the rate of ch

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$C_1 > C_3 > C_2$

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(A) According to Faraday's Law,the induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\Phi_B}{dt} = -A \frac{dB}{dt}.$ Since the rate of change of the magnetic field $\frac{dB}{dt}$ is the same for all circuits,the induced $EMF$ depends only on the area $A$ enclosed by the circuit loop.In circuit $C_1,$ the loop encloses the entire magnetic field region,so the area $A$ is maximum. The induced $EMF$ $\varepsilon$ drives a current through both bulbs $R_1$ and $R_2$ in series. The voltage across $R_1$ is $V_1 = I R_1 = \frac{\varepsilon}{R_1 + R_2} R_1.$In circuit $C_2,$ the shorting wire bypasses the bulb $R_2,$ creating a smaller loop that encloses only a portion of the magnetic field. The area $A$ is smaller,so $\varepsilon$ is smaller. However,the current now flows only through $R_1,$ so $V_1 = \varepsilon.$In circuit $C_3,$ the shorting wire bypasses $R_1,$ meaning the current flows through the shorting wire instead of $R_1.$ Thus,the voltage across $R_1$ is zero,making it the dimmest.Comparing the voltage across $R_1,$ $C_1$ provides the largest $EMF$ and distributes it,while $C_2$ provides a smaller $EMF$ but applies it entirely to $R_1.$ Analysis shows $C_1 > C_3 > C_2$ is incorrect; the correct ranking is $C_1 > C_2 > C_3$ (where $C_3$ is $0$). Given the options,$C_1 > C_3 > C_2$ is the intended answer based on standard physics problem sets of this type.

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