When a coil is placed in a time dependent magnetic field the power dissipated in it is P

Question

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 coil 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 ______. [2026]

Smart Achievers · Accepted Answer

(1)

$ϕ = B N A$

$Induced emf$ $| ε |$ $= N A \frac{d B}{d t}$

$Power dissipated: P = \frac{ε^{2}}{R}$

$Resistance of coil: R = ρ \frac{l}{π r^{2}}$

$Now: l = N (2 π R_{coil})$

$Also A = π R_{coil}^{2} \Rightarrow R_{coil} = \sqrt{\frac{A}{π}}$

$Thus R = \frac{ρ N (2 π)}{π r^{2}} \sqrt{\frac{A}{π}}$

$P = \frac{N^{2} A^{2} {(\frac{d B}{d t})}^{2}}{\frac{2 ρ N}{r^{2}} \sqrt{\frac{A}{π}}} \Rightarrow P \propto N A^{3 / 2} r^{2}$ $. . . (i)$

$P' \propto (2 N) {(2 A)}^{3 / 2} {(3 r)}^{2}$ $. . . (i i)$

$Dividing Eq. (ii)/Eq. (i)$

$\frac{P'}{P} = 36 \sqrt{2} \Rightarrow P' = (\sqrt{2}) (36) P$

$Thus α = 36 .$

Solution

1 $36$

2 $128 \sqrt{2}$

3 $16$

4 $64$

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Solution

1 36

2 1282

3 16

4 64

Ans. (1) ϕ=BNA Induced emf |ε|=NAdBdt Power dissipated: P=ε2R Resistance of coil: R=ρlπr2 Now: l=N(2πRcoil) Also A=πRcoil2⇒Rcoil=Aπ Thus R=ρN(2π)πr2Aπ P=N2A2(dBdt)22ρNr2Aπ⇒P∝NA3/2r2 ...(i) P'∝(2N)(2A)3/2(3r)2 ...(ii) Dividing Eq. (ii)/Eq. (i) P'P=362⇒P'=(2)(36)P Thus α=36.

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1 $36$

2 $128 \sqrt{2}$

3 $16$

4 $64$