The plane of a circular coil is at right angles to the magnetic meridian. If the number of turns in the coil equals 2, radius of the coil equals 0.078 m and current 2.8 A, then calculate the net magnetic field at the centre for clockwise and anticlockwise currents.
Given : `B_(u)=4 times 10^(-5)T.`

Explanation

Magnetic field at the cenre `=(mu_(0)/(4pi))((2pini)/r)tesla`
i.e., `" "B_(C)=((10^(-7))(2 times 3.142 times 2 times 2.8))/(0.078)`
`" "B_(C)=4.51 times 10^(-5)T.`
Case (i) : For anti clockwise current
`" "B_(R)=B+B_(H)`
`" "B_(R)=(4.51+4) times 10^(-5)T=8.51 times 10^(-5)T`
Case (ii) : For clockwise current
`" "B_(R)=B-B_(H)`
`" "=(4.51-4) times 10^(-5)`
`" "=0.51 times 10^(-5)T`