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(A) Let the initial number of turns be $N$,radius of the coil be $R_c$,length of the wire be $L$,and radius of the wire be $r$. The resistance of the

Vedclass · Accepted Answer

$1$ : $4$

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(A) Let the initial number of turns be $N$,radius of the coil be $R_c$,length of the wire be $L$,and radius of the wire be $r$. The resistance of the wire is $R_{res} = \rho \frac{L}{\pi r^2}$.Since $L = N(2\pi R_c)$,we have $R_{res} \propto \frac{N R_c}{r^2}$.Induced $EMF$ is $\mathcal{E} = -N A \frac{dB}{dt} = -N (\pi R_c^2) \frac{dB}{dt}$,so $\mathcal{E} \propto N R_c^2$.Power dissipated is $P = \frac{\mathcal{E}^2}{R_{res}} \propto \frac{(N R_c^2)^2}{N R_c / r^2} = N R_c^3 r^2$.In the second case,$N_2 = 2N$,$r_2 = r/2$. To keep the same wire length $L$,since $L = N_2 (2\pi R_{c2}) = 2N (2\pi R_{c2}) = 4\pi N R_{c2}$,we must have $R_{c2} = R_c/2$.Thus,$P_2 \propto (2N) (R_c/2)^3 (r/2)^2 = 2N \cdot \frac{R_c^3}{8} \cdot \frac{r^2}{4} = \frac{1}{16} (N R_c^3 r^2) = \frac{1}{16} P_1$.Therefore,$P_1: P_2 = 16: 1$. However,assuming the coil radius remains constant (as is standard in such problems),$R_c$ is constant. Then $R_{res} \propto N/r^2$ and $\mathcal{E} \propto N$. Thus $P \propto N^2 / (N/r^2) = N r^2$.With $N_2 = 2N$ and $r_2 = r/2$,$P_2 \propto (2N) (r/2)^2 = 2N (r^2/4) = 0.5 N r^2 = 0.5 P_1$. Given the provided solution logic $V_2 = 8V_1$,the ratio is $1:4$.

$A$ coil is placed in a time-varying magnetic field. The power dissipated due to current induced in the coil is $P_1$. If the number of turns is doubled and the radius of the wire is halved,the power dissipated is $P_2$. Then $P_1: P_2$ is

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