$A$ uniform conducting wire of length $12a$ and resistance $R$ is wound up as a current-carrying coil in the shape of:
$(i)$ an equilateral triangle of side $a$;
$(ii)$ a square of side $a$;
$(iii)$ a regular hexagon of side $a$.
The coil is connected to a voltage source $V_{0}$. Find the magnetic moment of the coils in each case.

(N/A) The magnetic dipole moment of a coil is given by $m = nIA$,where $n$ is the number of turns,$I$ is the current,and $A$ is the area of the loop. The current is $I = V_{0}/R$.
$(i)$ For an equilateral triangle of side $a$:
Perimeter of one turn $= 3a$.
Number of turns $n = 12a / 3a = 4$.
Area $A = \frac{\sqrt{3}}{4} a^2$.
Magnetic moment $m_1 = nIA = 4 \left( \frac{V_0}{R} \right) \left( \frac{\sqrt{3}}{4} a^2 \right) = \frac{\sqrt{3} V_0 a^2}{R}$.
$(ii)$ For a square of side $a$:
Perimeter of one turn $= 4a$.
Number of turns $n = 12a / 4a = 3$.
Area $A = a^2$.
Magnetic moment $m_2 = nIA = 3 \left( \frac{V_0}{R} \right) a^2 = \frac{3 V_0 a^2}{R}$.
$(iii)$ For a regular hexagon of side $a$:
Perimeter of one turn $= 6a$.
Number of turns $n = 12a / 6a = 2$.
Area $A = 6 \times \left( \frac{\sqrt{3}}{4} a^2 \right) = \frac{3\sqrt{3}}{2} a^2$.
Magnetic moment $m_3 = nIA = 2 \left( \frac{V_0}{R} \right) \left( \frac{3\sqrt{3}}{2} a^2 \right) = \frac{3\sqrt{3} V_0 a^2}{R}$.