Q.A uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current carrying coil in the shape of

Concept: magnetic moment of a coil, $m=NIA$.

The same wire (length $12a$, resistance $R$) is used for every shape, so across $V_0$ the current is the same throughout:

The number of turns is (total length)$\div$(perimeter of one turn), and each shape has its own area:

• Triangle ($3a$ perimeter): $N=\dfrac{12a}{3a}=4$, $A=\dfrac{\sqrt3}{4}a^2$

• Square ($4a$ perimeter): $N=\dfrac{12a}{4a}=3$, $A=a^2$ $m_\square=3\cdot\frac{V_0}{R}\cdot a^2=\frac{3V_0 a^2}{R}.$ …

Principle

A planar coil of $N$ turns carrying current $I$ and enclosing area $A$ per turn has magnetic moment

The wire length $12a$ is fixed, so a smaller perimeter per turn means more turns but less area — we must track both.

Step 1 — the current

The wire's resistance is $R$ regardless of how it is wound (resistance depends only on length, cross-section and resistivity). Across $V_0$:

the same for all three coils.

Step 2 — turns and area for each shape

Equilateral triangle, side $a$: perimeter $3a$, so $N=\dfrac{12a}{3a}=4$; area $A=\dfrac{\sqrt3}{4}a^2$.

Square, side $a$: perimeter $4a$, so $N=\dfrac{12a}{4a}=3$; area $A=a^2$.

$m_\square=3\cdot\frac{V_0}{R}\cdot a^2=\frac{3V_0 a^2}{R}.$ …

Method: Magnetic Moment of Coils Wound From a Fixed Length of Wire

This method applies whenever the SAME piece of wire (fixed total length) is wound into different shapes and connected to the same source, and you must compare the resulting magnetic moments.

Steps

Step 1: Recognise that the current is the same for every shape

Electrical resistance depends only on the wire's length, cross-section, and resistivity — not on how it is bent or coiled. So the same wire, of the same total length, has the same resistance $R$ in every configuration, and across the same source $V_0$ it carries the same current:

Compute this once; it applies to every shape.

Step 2: Find the number of turns for each shape

With a fixed total wire length $\ell_{\text{total}}$ and a shape of perimeter $p$,

A shape with a smaller perimeter per turn gives MORE turns from the same wire — but as Step 3 shows, more turns doesn't automatically mean more moment.

Step 3: Find the enclosed area per turn

Use the standard area formula for the shape (equilateral triangle $\tfrac{\sqrt3}{4}a^2$, square $a^2$, regular hexagon $\tfrac{3\sqrt3}{2}a^2$, etc., for side $a$). A shape that packs more area into the same perimeter (a hexagon vs a triangle, approaching a circle) encloses more area per turn even though it needs fewer turns.

Step 4: Combine into the magnetic moment …