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)`

Explanation

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`.