A coil of 200 turns and area 0 . 20 m 2 is rotated at half a revolution per second and is placed in uniform…

A coil of 200 turns and area 0.20 m2 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 2πβ volt. The value of β is ______.

Solution

Time taken for one complete revolution = time period =2 s.

Therefore, ω=2πT=π rad s-1

Now flux, ϕ=NABcos(ωt)

Therefore, induced EMF ε=-dϕdt=NABωsin(ωt)

The maximum voltage generated, εmax=NABω

=200×0.2×0.01×π

=4π10=2π5 volt

Hence, β=5.

Asked in: JEE Main 2024 (01 Feb Shift 2)

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