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Electromagnetic Induction

JEE Main level, single correct, medium

5 of 153

Question 5 of 153

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 _______.

Show answer and solution

Answer: (A) $36$

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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