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A cirular coil of wire cansiting of 100 turns ,each of readius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field barB at the centre of the coil ? |
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Answer» Solution :`n= 100 ,I=0.40 A, R= 8.0cm = 8.0 xx 10^(-2) m` `B = (mu_0nl )/(2r) = (4PI xx 10^(-7) xx 100 xx 0.4 )/(2xx8xx10^(-2))= PI xx 10^(-4) T` |
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