obtain the expression for magnetic field at axial point distance d from the center of a circular coil of radius a carrying current I . Also find the ratio of the magnitudes of magnetic field of this coil at the center and at axial point for which d=a√3

Dear Student,

Please find below the solution to the asked query:

The situation is shown in the figure below,

Here, consider a small element of length "ds" on the ring, If "dB' be the magnetic field induction due to this element at a point P as shown in the figure.
$dB = \frac{{\mu }_{0}I}{4\pi }\frac{ds}{r^2} = \frac{{\mu }_{0}I}{4\pi }\frac{ds}{\left(a^2+x^2\right)}$
Here, I = current passing through the loop, 'a' and 'x' are constants.
But here, for every part, there is an opposite part on the ring, so only their $\cos \theta$ components will get added.
So, $$\begin{gathered} B = \int dB \cos \theta = \frac{{\mu }_{0}I}{4\pi }\int \frac{ds \cos \theta }{\left(a^2+x^2\right)} = \frac{{\mu }_{0}I}{4\pi }\int \frac{ds}{\left(a^2+x^2\right)}\frac{a}{{\left(a^2+x^2\right)}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \Rightarrow B = \frac{{\mu }_{0}I a}{4\pi {\left(a^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\int }_0^{2\pi a} ds \\ \Rightarrow B= \frac{{\mu }_{0}I a^2}{2 {\left(a^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}} \end{gathered}$$

At the centre of the coil,
$B_0 = \frac{{\mu }_{0}I}{2a}$
and at a distance $x = a\sqrt{3}$ on the axial point,
$$\begin{gathered} B_1 = \frac{{\mu }_{0}I{a}^2}{2{\left(a^2+{\left(a\sqrt{3}\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{{\mu }_{0}I{a}^2}{2{\left(4a^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} = \frac{{\mu }_{0}I{a}^2}{16a^{{\displaystyle 3}}} \\ \Rightarrow B_1 = \frac{{\mu }_{0}I}{16a} \end{gathered}$$

Therefore, the ratio, 
$\frac{B_0}{B_1}=\frac{{\displaystyle \left(\frac{{\mu }_{0}I}{2a}\right)}}{\left(\frac{{\mu }_{0}I}{16a}\right)}=8$


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