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A circular coil of 10 turns and mean radius of 0.1 m is kept with its plane in the magnetic meridian. If a current of 2 A passes through it, calculate the resultant magnetic field at its centre. (B_(H)=4 times 10^(-5)T) |
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Answer» Solution :Magnetic FIELD at the centre, `""=(mu_(0)/(4pi))((2pini)/(R))tesla` `""B_(C)=(10^(-7) times 2 times 3.142 times 10.2)/(0.1) tesla` `""=12.57 times 10^(-5)T.` `""B_(R)=sqrt(B_(H)^(2)+B_(C)^(2))` `""=10^(-5)sqrt((4)^(2)+(12.57)^(2))` `""=10^(-5)sqrt(174.0)` `""B_(R)=13.19 times 10^(-5)T` Resultant magnetic field `=13.2 times 10^(-5)T`. 
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