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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 :Magnetic FIELD PRODUCED by a current carrying circular coil of radius R with N no. of turns wound tightly at the centre of coil is, `B=(mu_(0)NI)/(2R)` `thereforeB=((4pixx10^(-7))(100)(1))/((2)(0.1))` `thereforeB=6.28xx10^(-4)T` (Perpendicular to plane of the coil) |
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