Electromagnetic Induction Question 2
Question 2 - 2024 (01 Feb Shift 2)
A coil of 200 turns and area $0.20 \mathrm{~m}^{2}$ is rotated at half a revolution per second and is placed in uniform magnetic field of $0.01 \mathrm{~T}$ 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
Show Answer
Answer: (5)
Solution:
$\phi=\mathrm{NAB} \cos (\omega \mathrm{t})$
$\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{NAB} \omega \sin (\omega \mathrm{t})$
$\varepsilon_{\max }=\mathrm{NAB} \omega$
$=200 \times 0.2 \times 0.01 \times \pi$
$=\frac{4 \pi}{10}=\frac{2 \pi}{5}$ volt
