In figure bcdf is a circular coil of non-insulated thin uniform conductors, ab and de are very long straight parallel conductors, ab and de are very long straight parallel conductors, tangential to the coil at the points b and d. If the current `5A` enters the coil from a to b, find the magnetic field induction at O, the centre of the coil. The diameter of the coil is `10cm`.

Explanation

The current in the part bcd of the coil is equal to the current in the part `bfd` of the coil, which is equal to `2.5A`. Here,
`r=Ob=Oc=Od=Of=5cm=5xx10^-2m`.
Magnetic field induction at O due to current through circular coil will be zero because the magnetic field induction at O due to current through segment `bcd` of the coil is equal and opposite to that due to current through segment `bfd` of the coil.
Magnetic field induction at O due to current through long straight conductor ab is
`B_1=(mu_0)/(4pi)I/r(sin 90^@+sin 0^@)`
`=(mu_0)/(4pi)I/r=10^-7xx(5)/(5xx10^-2)=10^-5T`
outward normally to the plane of paper.
Magnetic field induction at O due to current through long staight conductor `de` is
`B_2=(mu_0)/(4pi)I/r(sin 90^@+sin0^@)`
`=(mu_0)/(4pi)I/r=10^-7xx(5)/(5xx10^-2)=10^-5T`
outwards normally to the plane of paper
`:.` Total magnetic field induction at O
`B=B_1+B_2=10^-5+10^-5=2xx10^-5T`
normal to plane of paper outwards.