Question

A coil of radius $10cm$ , carrying a current of $1A$ produces a magnetic field of induction $6.284\times {10}^{-3}Wb/m^2$ at its centre. What is the number of turns of the coil?

Answer 1

Magnetic field is given by:

$B=\frac{{\mu }_{0}ni}{2r}$

$6.284\times {10}^{-3}=\frac{4\pi \times {10}^{-7}\times n\times 1}{2\times 0.1}$

$n=\frac{2\times 6.284\times {10}^{-4}}{4\pi \times {10}^{-7}}$

$n=1000$ turns.

Thus number of turns of coil is 1000.