Electromagnetic Induction Question 2
Question 2 - 2024 (01 Feb Shift 2)
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 $\frac{2 \pi}{\beta}$ volt. The value of $\beta$ is
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Answer: (5)
Solution:
$\phi=NAB \cos (\omega t)$
$\varepsilon=-\frac{d \phi}{dt}=NAB \omega \sin (\omega t)$
$\varepsilon _{\max }=NAB \omega$
$=200 \times 0.2 \times 0.01 \times \pi$
$=\frac{4 \pi}{10}=\frac{2 \pi}{5}$ volt
