A uniform conducting wire of length is 24$$a$$, and resistance $$R$$ is wound up as a current carrying coil in the shape of an equilateral triangle of side $$a$$ and then in the form of a square of side $$a$$. The coil is connected to a voltage source $$V_0$$. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is $$1 : \sqrt{y}$$ where $$y$$ is _________.

We recall the definition of magnetic moment of a current-carrying planar coil. For any coil, the magnetic moment $$m$$ is given by the formula

$$m = N\,I\,A,$$

where $$N$$ is the number of turns, $$I$$ is the current flowing through the wire and $$A$$ is the area enclosed by a single turn of the coil. This relation will be applied successively to the triangular and the square coils made from the same wire.

The total length of the conducting wire is specified to be $$24a$$. Its resistance is given as $$R$$, so when it is connected to the same voltage source of emf $$V_0$$ in either shape, the current through the wire will be identical in the two cases. Using Ohm’s law, we have

$$I = \dfrac{V_0}{R}.$$

Because $$I$$ is the same for both coils, the ratio of their magnetic moments will depend only on the respective products $$N\,A$$. Let us therefore determine $$N$$ and $$A$$ for each geometry separately.

Triangular coil  The shape is an equilateral triangle with side length $$a$$. The perimeter of one turn is $$3a$$, so the number of turns that can be formed from a total length $$24a$$ is

$$N_{\triangle}= \dfrac{24a}{3a}=8.$$

The area of a single equilateral triangle of side $$a$$ is obtained from the standard formula

$$A_{\triangle}= \dfrac{\sqrt{3}}{4}\,a^{2}.$$

Square coil  The square has side length $$a$$. Its perimeter is $$4a$$, therefore the number of turns obtained from the same wire is

$$N_{\square}= \dfrac{24a}{4a}=6.$$

The area of a square of side $$a$$ is simply

$$A_{\square}= a^{2}.$$

Ratio of magnetic moments  Using $$m=NIA$$ for each coil and canceling the common factor $$I$$, we write

$$\dfrac{m_{\triangle}}{m_{\square}} \;=\;\dfrac{N_{\triangle} A_{\triangle}} {N_{\square} A_{\square}}.$$ Substituting the numerical values just obtained, we have

$$\dfrac{m_{\triangle}}{m_{\square}} = \dfrac{\,8 \times \dfrac{\sqrt{3}}{4}a^{2}\,} {\,6 \times a^{2}\,}.$$

Simplifying step by step,

$$\dfrac{m_{\triangle}}{m_{\square}} = \dfrac{8}{4}\;\times\;\dfrac{\sqrt{3}}{6} = 2 \times \dfrac{\sqrt{3}}{6} = \dfrac{\sqrt{3}}{3}.$$

The statement of the problem expresses this same ratio in the form

$$\dfrac{m_{\triangle}}{m_{\square}} = \dfrac{1}{\sqrt{y}}.$$

Accordingly, we equate

$$\dfrac{1}{\sqrt{y}} = \dfrac{\sqrt{3}}{3}.$$

Cross-multiplying,

$$\sqrt{y} = \dfrac{3}{\sqrt{3}} = \sqrt{3}.$$

Finally, squaring both sides gives

$$y = 3.$$

So, the answer is $$3$$.