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Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil? |
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Answer» Solution :Since the coil is tightly wound, we may take each circular ELEMENT to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is `B= (mu_0 NI)/(2R) = (4PI xx 10^(-7) xx 10^2 xx 1)/(2 xx 10^(-1) ) = 2pi xx 10^(-4) = 6.28 xx 10^(-4)T` |
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