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A current carrying coil (I = 5A, R = 10 cm.) having 50 number of turns find field at its centre-A. 1.57 mTB. 3.14 mTC. 1 mTD. 2 mT |
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Answer» Correct Answer - A Magnetic field at the centre of coil `B = (mu_(0)iN)/(2a)` `(4pixx10^(-7)xx5xx50)/(2xx10//100)=1.57xx10^(-3)T` =1.57mT. |
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