Physics - Magnetism Question with Solution | TestHub

Let the length of the wire be $L$.

For a circular coil of one turn, the radius $R_1$ is given by $L=2\pi R_1\Rightarrow R_1=\frac{L}{2\pi }$.

The magnetic field at the centre of the coil of one turn is $B_1=\frac{{\mu }_{0}I}{2R_1}$.

For a circular coil of $n$ turns, the radius $R_n$ is given by $L=n(2\pi R_n)\Rightarrow R_n=\frac{L}{2\pi n}=\frac{R_1}{n}$.

The magnetic field at the centre of the coil of $n$ turns is $B_n=\frac{{\mu }_{0}nI}{2R_n}$.

Substituting $R_n=\frac{R_1}{n}$, we get $B_n=\frac{{\mu }_{0}nI}{2\left(\frac{R_1}{n}\right)}=n^2\left(\frac{{\mu }_{0}I}{2R_1}\right)=n^2{B}_1$.

The ratio of the magnetic field for one turn to that of $n$ turns is $\frac{B_1}{B_n}=\frac{1}{n^2}$.

Answer: $1:n^2$