JEE Main 2026 — Electromagnetic Induction Question with Solution

The induced emf in the coil is given by Faraday's law:

$E=NA\frac{dB}{dt}$

The resistance of the coil is $R=\rho \frac{l}{a}$, where $l$ is the total length of the wire and $a$ is its cross-sectional area.

The length of the wire is $l=N(2\pi R_c)$, where $R_c$ is the radius of the coil. Since the area of the coil is $A=\pi R_c^2$, we have $R_c=\sqrt{\frac{A}{\pi }}$.

Thus, $l=2N\sqrt{\pi A}$.

The cross-sectional area of the wire is $a=\pi r^2$.

Therefore, the resistance is:

$R=\rho \frac{2N\sqrt{\pi A}}{\pi r^2}\propto \frac{N\sqrt{A}}{r^2}$

The power dissipated in the coil is:

$P=\frac{E^2}{R}\propto \frac{(NA{)}^2}{\frac{N\sqrt{A}}{r^2}}=N{A}^{3/2}{r}^2$

For the second coil, the parameters are $N^{\prime }=2N$, $A^{\prime }=2A$, and $r^{\prime }=3r$. The new power dissipated is:

$P^{\prime }=P\left(\frac{2N}{N}\right){\left(\frac{2A}{A}\right)}^{3/2}{\left(\frac{3r}{r}\right)}^2$

$P^{\prime }=P(2)(2\sqrt{2})(9)=36\sqrt{2}P$

Given that $P^{\prime }=\sqrt{2}\alpha P$, we can compare the expressions to find:

$\sqrt{2}\alpha =36\sqrt{2}\Rightarrow \alpha =36$

Answer: $36$