A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil

Q: Why is the current $I$ the same in both cases?

The problem states that the coil is connected to a voltage source $V_0$. Since the same wire is used, its total resistance $R$ remains unchanged. By Ohm's law, $I = \frac{V_0}{R}$, so the current is identical in both configurations.

Q: How did we get the number of turns?

The total length of the wire is $24a$. The perimeter of one turn of the triangle is $3a$, so the number of turns is $\frac{24a}{3a} = 8$. Similarly, the perimeter of one turn of the square is $4a$, giving $\frac{24a}{4a} = 6$ turns.

Q: What is the formula for the area of an equilateral triangle?

The area of an equilateral triangle with side length $a$ is $\frac{\sqrt{3}}{4}a^2$.

Q: Why do we use $M = nIA$ instead of just $M = IA$?

The formula $M = IA$ is for a single loop. When a wire is wound into a coil with $n$ turns, each turn contributes to the magnetic moment, so the total magnetic moment becomes $n$ times that of a single loop, i.e., $M = nIA$.

Q: Can we assume the wire's thickness affects the area?

In such textbook problems, the wire is considered 'thin', meaning its thickness is negligible compared to the dimensions of the coil. Thus, we use the side length $a$ directly to calculate the area.

Question

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 $1: \sqrt{y}$. The value of where $y$ is ......... .

Sigma Percentile · Accepted Answer

### The Magic of a Single Wire

Imagine you have a long, uniform conducting wire of length $l = 24a$ and resistance $R$. You are tasked with winding this wire into two different shapes: first, an equilateral triangle of side $a$, and second, a square of side $a$. Both coils are then connected to the same voltage source $V_0$. Our goal is to find the ratio of their magnetic moments.

This problem is a beautiful exercise in geometry and electromagnetism. Let's break it down step by step.

### Analyzing the Setup

The most crucial realization here is the **conservation of length**. The total length of the wire remains $24a$ regardless of the shape it takes. This allows us to determine the number of turns in each coil.

For the equilateral triangle, the perimeter of a single turn is $3a$. Therefore, the total number of turns $n_T$ is:
$$n_T = \frac{24a}{3a} = 8$$

Similarly, for the square, the perimeter of a single turn is $4a$. The total number of turns $n_S$ is:
$$n_S = \frac{24a}{4a} = 6$$

### The Master Equation

The magnetic moment $M$ of a current-carrying coil is given by the product of the number of turns $n$, the current $I$, and the area $A$:
$$M = nIA$$

Since both coils are made from the same wire, their total resistance $R$ is identical. When connected to the same voltage source $V_0$, Ohm's law tells us that the current $I = \frac{V_0}{R}$ will be exactly the same for both coils. This is a massive simplification because when we take the ratio of their magnetic moments, the current $I$ will simply cancel out!

### Final Calculation

Let's set up the ratio of the magnetic moment of the triangular coil $M_T$ to that of the square coil $M_S$:
$$\frac{M_T}{M_S} = \frac{n_T I A_T}{n_S I A_S} = \frac{n_T A_T}{n_S A_S}$$

Now, we need the areas of both shapes. The area of an equilateral triangle of side $a$ is:
$$A_T = \frac{\sqrt{3}}{4}a^2$$

The area of a square of side $a$ is:
$$A_S = a^2$$

Substituting these values into our ratio equation:
$$\frac{M_T}{M_S} = \frac{8 	imes \left(\frac{\sqrt{3}}{4}a^2ight)}{6 	imes a^2}$$

The $a^2$ terms cancel out beautifully:
$$\frac{M_T}{M_S} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$

The problem states that this ratio is equal to $1 : \sqrt{y}$. By comparing our result with the given expression:
$$\frac{1}{\sqrt{y}} = \frac{1}{\sqrt{3}}$$

Squaring both sides, we find our final answer:
**$$y = 3$$**

This elegant problem demonstrates how physical constraints (like a fixed wire length) dictate the geometric properties (number of turns) and ultimately govern the electromagnetic behavior (magnetic moment) of a system.

Enter Numerical Value:

Visualized Solution

$l = 24 a$

$n = \frac{l}{perimeter}$

$M = n I A$

$\frac{M _{T}}{M _{S}}$

$A_{T}$ \text{ and } $A_{S}$

$\frac{M _{T}}{M _{S}} = \frac{1}{3}$

$y = 3$

\text{Conclusion}

The Sigma Insight: Magnetic Moment of Current Loop

The Magic of a Single Wire

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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