A uniform conducting wire of length is $24a$, 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 ______.
Numerical value type. Enter your answer.
Answer: 3
The resistance (and hence the current) is the same in both cases.
Triangle: $\dfrac{24a}{3a} = 8$ turns of area $\dfrac{\sqrt{3}}{4}a^2$, so $m_1 = 8I\cdot\dfrac{\sqrt{3}}{4}a^2 = 2\sqrt{3}Ia^2$.
Square: $\dfrac{24a}{4a} = 6$ turns of area $a^2$, so $m_2 = 6Ia^2$.
$$\frac{m_1}{m_2} = \frac{2\sqrt{3}}{6} = \frac{1}{\sqrt{3}} \Rightarrow y = 3$$
Solution by Sreeraj P, M.Sc Physics