Q63 / 75 · JEE Main 2026 (02 April Shift 1) · Physics
When a coil is placed in a time dependent magnetic field the power dissipated in it is $P$. The number of turns, area of the coil and radius of the coil wire are $N$, $A$ and $r$ respectively. For a second coils number of turns, area of the coil and radius of the coil wire are $2N$, $2A$ and $3r$ respectively. When the first coil is replaced with second coil the power dissipated in it is $\sqrt{2}\,\alpha P$. The value of $\alpha$ is _______.
- Correct answerA$36$
- B$128\sqrt{2}$
- C$16$
- D$64$
Solution
The induced emf in the coil is given by Faraday's law:
$E = N A \dfrac{dB}{dt}$
The resistance of the coil is $R = \rho \dfrac{l}{a}$, 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 $R_c = \sqrt{\dfrac{A}{\pi}}$.
Thus, $l = 2N\sqrt{\pi A}$.
The cross-sectional area of the wire is $a = \pi r^2$.
Therefore, the resistance is:
$R = \rho \dfrac{2N\sqrt{\pi A}}{\pi r^2} \propto \dfrac{N\sqrt{A}}{r^2}$
The power dissipated in the coil is:
$P = \dfrac{E^2}{R} \propto \dfrac{(N A)^2}{\dfrac{N\sqrt{A}}{r^2}} = N A^{3/2} r^2$
For the second coil, the parameters are $N' = 2N$, $A' = 2A$, and $r' = 3r$. The new power dissipated is:
$P' = P \left(\dfrac{2N}{N}\right) \left(\dfrac{2A}{A}\right)^{3/2} \left(\dfrac{3r}{r}\right)^2$
$P' = P (2) (2\sqrt{2}) (9) = 36\sqrt{2} P$
Given that $P' = \sqrt{2}\alpha P$, we can compare the expressions to find:
$\sqrt{2}\alpha = 36\sqrt{2} \Rightarrow \alpha = 36$
Answer: $36$
$E = N A \dfrac{dB}{dt}$
The resistance of the coil is $R = \rho \dfrac{l}{a}$, 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 $R_c = \sqrt{\dfrac{A}{\pi}}$.
Thus, $l = 2N\sqrt{\pi A}$.
The cross-sectional area of the wire is $a = \pi r^2$.
Therefore, the resistance is:
$R = \rho \dfrac{2N\sqrt{\pi A}}{\pi r^2} \propto \dfrac{N\sqrt{A}}{r^2}$
The power dissipated in the coil is:
$P = \dfrac{E^2}{R} \propto \dfrac{(N A)^2}{\dfrac{N\sqrt{A}}{r^2}} = N A^{3/2} r^2$
For the second coil, the parameters are $N' = 2N$, $A' = 2A$, and $r' = 3r$. The new power dissipated is:
$P' = P \left(\dfrac{2N}{N}\right) \left(\dfrac{2A}{A}\right)^{3/2} \left(\dfrac{3r}{r}\right)^2$
$P' = P (2) (2\sqrt{2}) (9) = 36\sqrt{2} P$
Given that $P' = \sqrt{2}\alpha P$, we can compare the expressions to find:
$\sqrt{2}\alpha = 36\sqrt{2} \Rightarrow \alpha = 36$
Answer: $36$
Hence, the correct answer is option: 'A'.