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A circular coil of wire consisting of 100 turns, each of radius 8.0 cm, carries a current of 0.40 A. What is the magnitude field vecB at the centre of the coil? |
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Answer» Solution :Here `N = 100 , R = 8.0 cm = 8 xx 10^(-2) m, l = 0.40 A` `:.` Magnitude of the MAGNETIC FIELD at the centre of coil `B = (mu_0 NI)/(2R) = (4 pi xx 10^(-7) xx 100 xx 0.4)/(2 xx 8 xx 10^(-2)) = pi xx 10^(-4) T " or " 3.14 xx 10^(-4) T`. |
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