A closely wound flat circular coil of 25 turns of wire has diameter of 10 cm which carries current of 4 A. The flux density at the centre of a coil will be

The magnetic field at the centre of the coil,

$\mathrm{B}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2\mathrm{\pi Ni}}{\mathrm{a}}$

where N is number of turns, i is current and a is radius of coil.

$\mathrm{Given},\mathrm{N}=25,\mathrm{D}=10\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$

Radius of the coil,

$\mathrm{a}=\frac{\mathrm{D}}{2}=\frac{10\times {10}^{-2}}{2}\mathrm{m}=0.05\mathrm{m}$

And current i = 4 A.  Putting these values in the expression for B , we get B=1.256 mT