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InterviewSolution
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A coil of 200 turns carries a current of 4 A. If the magnetic flux through the coil is 6 x 10-5 Wb, find the magnetic energy stored in the medium surrounding the coil. |
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Answer» Number of turns of the coil, N = 200 Current, I = 4 A Magnetic flux through the coil, Φ = 6 x 10-5 Wb Energy stored in the coil, U = \(\frac{1}{2}\) LI2 = \(\frac{1}{I^2}\) Self inductance of the coil, L = \(\frac{NΦ}{I}\) U = \(\frac{1}{2} \frac{NΦ}{I}\) x I2 = \(\frac{1}{2} \frac{NΦ}{I}\) = \(\frac{1}{2}\) x 200 x 6 x 10-5 x 4 U = 2400 x 10-5 = 0.024 J (or) joules. |
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