JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Electromagnetic Induction

In an ac generator, a rectangular coil of 100 turns each having area

14 \times 1 0^{- 2}

m² is rotated at 360 rev/min about an axis perpendicular to a uniform magnetic field of magnitude 3.0 T. The maximum value of the emf produced will be ______ V. (Take

π = \frac{22}{7}

)

Answer: 1584

Solution

The peak emf of a rotating coil is

ε_{0} = N A B ω

, with

N = 100

A = 14 \times 1 0^{- 2}

^{2}

B = 3.0

T and

ω = \frac{2 π \times 360}{60} = 12 π

rad s

^{- 1}

. Taking

π = \frac{22}{7}

, so that

ω = 12 \times \frac{22}{7}

, the maximum emf is

ε_{0} = 100 \times 0.14 \times 3 \times 12 \times \frac{22}{7} = 504 \times \frac{22}{7} = 1584

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer

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