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 voltag

Question

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 ______.

TestHub · Accepted Answer

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

Therefore, $ω = \frac{2 π}{T} = π rad s^{- 1}$

Now flux, $ϕ = NABcos (ω t)$

Therefore, induced EMF $ε = |- \frac{d ϕ}{dt}| = NAB ω \sin (ω t)$

The maximum voltage generated, $ε_{\max} = NAB ω$

$= 200 \times 0.2 \times 0.01 \times π$

$= \frac{4 π}{10} = \frac{2 π}{5}$ volt

Hence, $β = 5$ .

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