Question
The area of the coil must be$.......$
✓
Answer
Flux through a circular coil $\phi=\text{NBA}\cos\omega\text{t}$ Voltage required $\epsilon=\frac{-\text{d}\phi}{\text{dt}}$
$\Rightarrow9=\text{NBA}\omega\sin\omega\text{t}$
$9=\frac{8\times10^{-5}\times\text{A}\times30\times2\pi\times2000}{60}$
$\text{A}=\frac{9\times10^5}{50\times10^3}=18\text{m}^2$
$\Rightarrow9=\text{NBA}\omega\sin\omega\text{t}$
$9=\frac{8\times10^{-5}\times\text{A}\times30\times2\pi\times2000}{60}$
$\text{A}=\frac{9\times10^5}{50\times10^3}=18\text{m}^2$
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