JEE Main 2026 — Physics Electromagnetism

2026mcqhard

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 _______.

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$36$
B
$128\sqrt{2}$
C
$16$
D
$64$

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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: $36$

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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: $36$

JEE Main 2026 — Physics Electromagnetism

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JEE Main 2026 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism