Question

A closely wound coil has a radius of $6.00cm$ and carries a current of $2.50A$. How many turns it must have if, at a point on the coil axis $6.00cm$ from the centre of the coil,the magnetic field is $6.39\times {10}^{-4}T?$

Answer 1

The magnetic field at a distance $x$ from the center of circular coil is $B=\frac{{\mu }_{o}I{R}^2}{2{(R^2+x^2)}^{3/2}}$.
If the coil has $N$ turns , the magnetic field becomes $B=\frac{{\mu }_{o}NI{R}^2}{2(R^2+x^2{)}^{3/2}}$.

$N=\frac{2B(R^2+x^2{)}^{3/2}}{{\mu }_{0}I{R}^2}=\frac{2\times 6.39\times {10}^{-4}{\left[\right(0.06{)}^2+\left(0.06{)}^2\right]}^{3/2}}{4\pi \times {10}^{-7}\times 2.5\times (0.06{)}^2}=69$

thus n=69