Question

The radius of a coil of wire with $N$ turns is $0.22\text{m}$ and $3.5\text{A}$ current flows clockwise in the coil as shown. A long straight wire carrying a current $54\text{A}$ towards the left is located $0.05$ meters from the edge of the coil. The magnetic field at the centre of the coil is zero tesla. The number of turns $N$ in the coil are:

Answer 1

At centre of coil,
$B_{centre}=\frac{{\mu }_0{i}_1}{2\pi (d+r)}-\frac{N\times {\mu }_0{i}_2}{2r}$
Given $B_{centre}=0$
$\frac{{\mu }_0{i}_1}{2\pi (d+r)}=\frac{N\times {\mu }_0{i}_2}{2r}$
$\Rightarrow N=\frac{i_1}{i_2}.\frac{r}{\pi (d+r)}=\frac{54}{3.5}.\frac{0.22\times 7}{22\left(0.27\right)}=4$