JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Electromagnetic Induction

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A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:
(Assume the coil to be short circuited.)

(A)Halved
(B)Quadrupled
(C)The same
(D)Doubled✓ Correct

Answer: (D)

Solution

The power dissipated in the coil is given by:

P = \frac{ε ^{2}}{R} = \frac{( N A \frac{d B}{d t} ) ^{2} \times A _{C}}{ρ ℓ}

where

N

is the number of turns,

A

is the area,

A_{C}

is the cross-sectional area of the wire,

ρ

is resistivity, and

ℓ

is the length of wire.

When

N \to N /2

and wire radius is doubled (so

A_{C} \to 4 A_{C}

and resistance

R \to R /2

due to increased cross-section and halved length):

P^{'} = \frac{( \frac{N}{2} A \frac{d B}{d t} ) ^{2} \times 4 A _{C}}{ρ ℓ /2} = \frac{\frac{1}{4} ( N A \frac{d B}{d t} ) ^{2} \times 4 A _{C} \times 2}{ρ ℓ} = 2 P

Therefore, the power is doubled.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Electromagnetic Induction

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