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QJEEM23Ph5b8b: JEE Main 2023 Physics - Class 12 JEEM23Ph5b8b Solution: In an ac generator, a rectangular coil of 100 turns each having …

In an ac generator, a rectangular coil of 100 turns each having area 14×10−2 m214 \times 10^{-2} \mathrm{~m}^{2} is rotated at 360 rev/min360 ~\mathrm{rev} / \mathrm{min} about an axis perpendicular to a uniform magnetic field of magnitude 3.0 T3.0 \mathrm{~T} . The maximum value of the emf produced will be ________ VV .

(\left(\right. Take π=227)\left.\pi=\frac{22}{7}\right)

Step-by-Step Verified Solution

ϕ=B.A\phi=B.A

ϕ=BNAcos⁡ωt\phi=\mathrm{BNA}\cos\omega t

So, Emf=−dϕdt=NBAωsin⁡ωtEmf = {{ - d\phi } \over {dt}} = NBA\omega \sin \omega t

So maximum value of emf is

Emax⁡=NBAω{E_{\max }} = NBA\omega

=100×3×14×10−2×360×2π60=1584= 100 \times 3 \times 14 \times {10^{ - 2}} \times {{360 \times 2\pi } \over {60}} = 1584

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