Physics Q38 – JEE-MAIN 2026

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

2 N

2 A

and

3 r

respectively. When the first coil is replaced with second coil the power dissipated in it is

\sqrt{2} α P

. The value of

α

is _______.

Clue 1 of 3: Electromagnetic Induction and Power Dissipation

The power dissipated in a coil in a time-dependent magnetic field is due to the induced current, which is governed by Faraday's Law of Induction and Joule heating.

🧠 Full Solution Path

Step 1: Derive the expression for power dissipated✦ Active

The induced EMF in the coil is $E = N \frac{d Φ}{d t} = N \frac{d (B A)}{d t} = N A \frac{d B}{d t}$ . Let $\frac{d B}{d t} = k$ . So, $E = N A k$ .

The resistance of the coil wire is $R = ρ \frac{L}{A_{w i r e}}$ . The length of the wire is $L = N \times (2 π R_{c o i l})$ . Since $A = π R_{c o i l}^{2}$ , $R_{c o i l} = \sqrt{A / π}$ . Thus, $L = N \times 2 π \sqrt{A / π} = 2 N \sqrt{π A}$ .

The cross-sectional area of the wire is $A_{w i r e} = π r^{2}$ .

So, $R = ρ \frac{2 N \sqrt{π A}}{π r^{2}} = \frac{2 ρ N \sqrt{A}}{\sqrt{π} r^{2}}$ .

The power dissipated is $P = \frac{E^{2}}{R} = \frac{(N A k)^{2}}{\frac{2 ρ N \sqrt{A}}{\sqrt{π} r^{2}}} = \frac{N^{2} A^{2} k^{2} \sqrt{π} r^{2}}{2 ρ N \sqrt{A}} = \frac{N A^{3 / 2} k^{2} \sqrt{π} r^{2}}{2 ρ}$ .

Therefore, $P \propto N A^{3 / 2} r^{2}$ .

Step 2: Calculate the power for the second coil○ Expand

For the first coil, $P_{1} \propto N A^{3 / 2} r^{2}$ .

For the second coil, the number of turns is $N_{2} = 2 N$ , the area is $A_{2} = 2 A$ , and the radius of the wire is $r_{2} = 3 r$ .

Substituting these values into the proportionality:

P_{2} \propto (2 N) (2 A)^{3 / 2} (3 r)^{2}

P_{2} \propto (2) (2^{3 / 2}) (3^{2}) N A^{3 / 2} r^{2}

P_{2} \propto (2) (2 \sqrt{2}) (9) N A^{3 / 2} r^{2}

P_{2} \propto (36 \sqrt{2}) N A^{3 / 2} r^{2}

Step 3: Determine the value of

α

○ Expand

From the previous steps, we have the ratio:

\frac{P_{2}}{P_{1}} = \frac{(36 \sqrt{2}) N A^{3 / 2} r^{2}}{N A^{3 / 2} r^{2}} = 36 \sqrt{2}

The problem states that $P_{2} = \sqrt{2} α P_{1}$ . So, $\frac{P_{2}}{P_{1}} = \sqrt{2} α$ .

Equating the two expressions for the ratio:

36 \sqrt{2} = \sqrt{2} α

Solving for $α$ :

α = 36

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Physics Question 38 – JEE-MAIN 2026

Clue 1 of 3: Electromagnetic Induction and Power Dissipation

🧠 Full Solution Path