The magnetic field normal to the plane of a coil of \(N\) turns and radius \(r\) which carries a current…
The magnetic field normal to the plane of a coil of \(N\) turns and radius \(r\) which carries a current \(i\) is measured on the axis of the coil at a distance \(h\) from the centre of the coil. This is smaller than the field at the centre by the fraction,
\(\frac{3}{2} \cdot \frac{h^2}{r^2}\)
\(\frac{2}{3} \cdot \frac{h^2}{r^2}\)
\(\frac{3}{2} \cdot \frac{r^2}{h^2}\)
\(\frac{2}{3} \cdot \frac{r^2}{h^2}\)
Solution
The magnetic field normal to the plane of a coil of \(N\) turns and radius \(r\) having current \(i\) on the axis at a distance \(h\) from the centre is given by
\(\begin{aligned}
B_{\text {axis }} & =\frac{\mu_0 N i r^2}{2\left(r^2+h^2\right)^{\frac{3}{2}}} \\
& =\frac{\mu_0 N i r^2}{2\left(r^2\right)^{3 / 2}\left(1+\frac{h^2}{r^2}\right)^{\frac{3}{2}}}=\frac{\mu_0 N i r^2\left[1+\frac{h^2}{r^2}\right]^{\frac{3}{2}}}{2 r^3}
\end{aligned}\)
[From Binomial theorem, \((\mathbf{l}+x)^n=\mathbf{1}+n x\)]
\(B_{\text {axis }}=\frac{\mu_0 N i}{2 r}\left(1-\frac{3 h^2}{2 r^2}\right)\)...(i)
Magnetic field on the centre of a circular current carrying coil is given by
\(B_{\text {centre }}=\frac{\mu_0 N i}{2 r}\)...(ii)
From Eqs. (i) and (ii), we get
\(\begin{aligned}
& B_{\text {axis }}=B_{\text {centre }}\left(1-\frac{3 h^2}{2 h^2}\right) \\
& B_{\text {axis }}=B_{\text {centre }}-\frac{3 h^2}{2 r^2} \cdot B_{\text {centre }} \quad \ldots (ii)
\end{aligned}\)
From Eq. (iii), it is clear that the magnetic field on the axis is smaller than the field at the centre by a fraction \(\frac{3 h^2}{2 r^2}\).