A square shaped coil of area \(70\,cm ^2\) having \(600\) turns rotates in a magnetic field of \(0.4\,wbm ^{-2}\), about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes \(500\) revolution in a minute, the instantaneous emf when the plane of the coil is inclined at \(60^{\circ}\) with the field, will be \(..........V\). (Take \(\pi=\frac{22}{7}\) )

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Answer & Solution

Correct Answer

(D) \(44\)

Detailed explanation

\(\omega = 2\pi f = 2\pi \left(\frac{500}{60}\right) = \frac{50\pi}{3}\,rad/s\)

\(\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}\)

\(e = NBA\omega \sin\theta\)

\(e = 600 \times 0.4 \times (70 \times 10^{-4}) \times \frac{50\pi}{3} \times \sin(30^{\circ})\)

\(e = 600 \times 0.4 \times 0.0070 \times \frac{50\pi}{3} \times \frac{1}{2}\)

\(e = 14\pi\)

\(e = 14 \times \frac{22}{7} = 44\,V\)

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