A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. I

Q: Why does the magnetic moment change if the current and the wire are exactly the same?

Because the magnetic moment depends directly on the *area* enclosed by the loop ($m = IA$). Different geometric shapes with the exact same perimeter enclose different amounts of area.

Q: How do we know the length of the wire is constant?

The problem states 'this square loop is changed to a circular loop', implying the same physical wire is bent into a new shape without stretching or cutting it. Therefore, Perimeter of Square = Circumference of Circle.

Q: What is the formula for the magnetic moment of a coil with N turns?

The general formula is $m = NIA$, where $N$ is the number of turns. In this specific problem, it is a single loop, so $N=1$ for both the square and the circle.

Q: Does a circle always have a larger magnetic moment than a square for the same wire length?

Yes! According to a mathematical principle called the isoperimetric inequality, a circle encloses the maximum possible area for any given perimeter. Thus, it will always maximize the magnetic moment.

Q: What if the wire was reshaped into an equilateral triangle instead?

You would follow the exact same logic! Equate the perimeters: $4a = 3s$ (where $s$ is the triangle's side). Find the new area $\frac{\sqrt{3}}{4}s^2$, and calculate the new magnetic moment. You would find it is actually smaller than both the square and the circle!

Question

A square loop is carrying a steady current $I$ and the magnitude of its magnetic dipole moment is $m$. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop (in A-m) will be

Options:

(A) $\frac{4m}{\pi}$

(B) $\frac{3m}{\pi}$

(C) $\frac{2m}{\pi}$

(D) $\frac{m}{\pi}$

Sigma Percentile · Accepted Answer

### The Shape of Magnetism: From Square to Circle

Imagine you have a piece of wire carrying a steady current $I$. When you bend this wire into a closed loop, it acts like a tiny magnet, possessing what we call a **magnetic dipole moment**. The magnitude of this magnetic moment is remarkably simple to calculate: it is just the product of the current and the area enclosed by the loop.

$$m = I \cdot A$$

### Analyzing the Setup

Let's start with our initial configuration: a square loop. If we assume the side length of this square is $a$, the area it encloses is simply $a^2$. Therefore, the initial magnetic dipole moment $m$ is given by:

$$m = I \cdot a^2$$

Now, the problem asks us to reshape this exact same wire into a circular loop. The current $I$ remains unchanged. But what about the area? Since the shape has changed, the area will change, and consequently, the magnetic moment will change too.

### The Master Equation: Conservation of Length

The crucial physical constraint here is that the total length of the wire remains constant. This means the perimeter of the square must perfectly equal the circumference of the new circle. Let the radius of the new circle be $r$.

$$4a = 2\pi r$$

From this, we can easily express the new radius $r$ in terms of our original square's side $a$:

$$r = \frac{4a}{2\pi} = \frac{2a}{\pi}$$

### Final Calculation

With the radius known, we can now find the area of the circular loop, $A' = \pi r^2$. Let's calculate the new magnetic moment $m'$:

$$m' = I \cdot A' = I \cdot \pi r^2$$

Substituting our expression for $r$:

$$m' = I \cdot \pi \left(\frac{2a}{\pi}\right)^2$$

$$m' = I \cdot \pi \cdot \frac{4a^2}{\pi^2}$$

$$m' = I \cdot \frac{4a^2}{\pi}$$

Now, look closely at this final expression. We can factor out the term $I \cdot a^2$.

$$m' = \frac{4}{\pi} (I \cdot a^2)$$

But wait, $I \cdot a^2$ is exactly our original magnetic moment $m$! Substituting this back in, we arrive at our final, elegant result:

$$m' = \frac{4m}{\pi}$$

This result beautifully illustrates the **isoperimetric inequality**: for a given perimeter, a circle encloses the maximum possible area. Since the magnetic moment is directly proportional to the area, reshaping any loop into a circle will always maximize its magnetic dipole moment!

Select Answer:

Visualized Solution

$m = I \cdot A$

$Perimeter Conservation$

$r = \frac{2 a}{π}$

$m^{'} = I \cdot A^{'}$

$m^{'} = \frac{4 m}{π}$

$Conclusion$

The Sigma Insight: Magnetic Moment of Current Loop

The Shape of Magnetism

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=I⋅A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Perimeter Conservation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=π2a​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m′=I⋅A′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m′=π4m​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion