Question

Establish the formula for intensity of magnetic field at the centre of a current carrying circular coil.

Answer 1

Consider a circular coil of radius r with n turns carrying i. The intensity of magnetic field at its centre O due to an small part AB$=\Delta l$ will be.
$AB=\frac{{\mu }_o}{4\pi }\frac{i\Delta lsin{90}^o}{r^3}$
[Here $\theta ={90}^o$ because the radius is normal to the circumference.]
The intensity of magnetic field at the centre O due to the entire coil,
$\sum AB=\sum \frac{{\mu }_{o}i\Delta l}{4\pi r^2}$ $[\because sin{90}^o=1]$
$B=\frac{{\mu }_{o}i}{4\pi r^2}\sum \Delta l=\frac{{\mu }_o}{4\pi }\frac{i}{r^2}\times 2\pi rn$
Here $\sum \Delta B=B$ and $\sum \Delta l=2\pi rn=$circumference of the coil
$\therefore B=\frac{{\mu }_o}{4\pi }\frac{2\pi ni}{r}$
$B=\frac{{\mu }_{o}ni}{2r}N/Am$
This is the required expression.
If the direction of current in the coil is anticlockwise, the direction of magnetic field will be normal to the plane of paper upwards and if current in the coil is clockwise, the direction of magnetic field is normal to the plane to paper downwards.