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JEE Main20241 Feb 2024Evening ShiftPhysicsElectromagnetic InductionActual

A coil of 200 turns and area 0 . 20 m 2 is rotated at half a revolution per second and is placed in uniform magnetic field of 0 . 01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2 π β volt. The value of β is ______.

Correct answer

0

Step-by-step solution

Time taken for one complete revolution = time period = 2 s . Therefore, ω = 2 π T = π rad s - 1 Now flux, ϕ = NABcos ( ω t ) Therefore, induced EMF ε = - d ϕ dt = NAB ω sin ( ω t ) The maximum voltage generated, ε max = NAB ω = 200 × 0 . 2 × 0 . 01 × π = 4 π 10 = 2 π 5 volt Hence, β = 5 .

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