JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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 \(\pi=\frac{22}{7}\) )
- A \(40\)
- B \(42\)
- C \(46\)
- D \(44\)
Answer & Solution
Correct Answer
(D) \(44\)
Step-by-step Solution
Detailed explanation
\(\omega = 2\pi f = 2\pi \left(\frac{500}{60}\right) = \frac{50\pi}{3}\,rad/s\) \(\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}\) \(e = NBA\omega \sin\theta\) \(e = 600 \times 0.4 \times (70 \times 10^{-4}) \times \frac{50\pi}{3} \times \sin(30^{\circ})\)…
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