A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s ⁻¹ in a uniform horizontal magnetic field of 3.0 × 10 - 2 T. The maximum emf induced the coil will be................. × 10 - 2 volt (rounded off to the nearest integer) Numerical answer type. Work it out, then open the answer.

Answer: 60. Peak emf: e₀ = NBAω with A = π r² = π(0.08)² = 0.0201 m². e₀ = 20 × 3×10⁻² × 0.0201 × 50 e₀ ≈ 0.603 V = 60 × 10⁻² V Answer: 60

A circular coil of radius 8.0 cm and 20 turns is rotated about its… | JEE Main 2021 (26 Aug, Shift 2)

Q25JEE Main 2021 · 26 AUG · SHIFT 2Easy

Rotating Conductors & AC Generator

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s ${}^{-1}$ in a uniform horizontal magnetic field of 3.0 $\times$ 10 $-$ 2 T. The maximum emf induced the coil will be ................. $\times$ 10 $-$ 2 volt (rounded off to the nearest integer)

Numerical answer type. Work it out, then open the answer.

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Answer: 60

Peak emf: $e_0 = NBA\omega$ with $A = \pi r^2 = \pi(0.08)^2 = 0.0201\text{ m}^2$ .
$e_0 = 20 \times 3\times10^{-2} \times 0.0201 \times 50$
$e_0 \approx 0.603\text{ V} = 60 \times 10^{-2}\text{ V}$
Answer: $60$

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