A conductor 10 cm long is moves with a speed 1 m/s perpendicular to a field of strength 1000 A/m. The emf induced in the conductor is (Given: μ₀ = 4π × 10⁻⁷ Wb/Am) (a) π mV (b) 2π mV (c) 40πμ V (d) 4πμ V

Answer: (c). Field strength H is given, so B = μ₀ H = 4π×10⁻⁷×1000 = 4π×10⁻⁴ T. ε = Blv = 4π×10⁻⁴×0.1×1 = 4π×10⁻⁵ V = 40π μ V. So the answer is (c).

A conductor 10 cm long is moves with a speed 1 m/s perpendicular to a… | MHT CET 2023 (13 May, Shift 2)

Q23MHT CET 2023 · 13 MAY · SHIFT 2Easy

Motional EMF

A conductor $10\text{ cm}$ long is moves with a speed $1\text{ m/s}$ perpendicular to a field of strength $1000\text{ A/m}$ . The emf induced in the conductor is (Given: $\mu_0 = 4\pi \times 10^{-7}\text{ Wb/Am}$ )

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Answer: (c)

Field strength $H$ is given, so $B = \mu_0 H = 4\pi\times10^{-7}\times1000 = 4\pi\times10^{-4}$ T.
$\varepsilon = Blv = 4\pi\times10^{-4}\times0.1\times1 = 4\pi\times10^{-5}$ V
$= 40\pi\ \mu$ V. So the answer is (c).

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Motional EMF question from MHT CET 2023 (13 May, Shift 2)

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