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’. The magnetic dipole moments of the coil in each case respectively are :

The correct option is √ 3 Ia² and 3 Ia².

Step 1: Understand the formula for the magnetic dipole moment
The magnetic dipole moment ($M$) of a current-carrying coil with $N$ turns, carrying current $I$, and enclosing an area $A$ is given by:
$M=N·I·A$
The total length of the uniform conducting wire is $L=12a$.

Step 2: Calculate the magnetic dipole moment for the equilateral triangular coil
For an equilateral triangle of side $a$:
The perimeter of a single turn is $3a$.
The number of turns $N_1$ that can be made from the wire of length $12a$ is:
$N_1=\frac{12a}{3a}=4$

The area $A_1$ of an equilateral triangle of side $a$ is:
$A_1=\frac{\sqrt{3}}{4}{a}^2$

Using the formula for magnetic moment, we get:
$M_1=N_1·I·A_1=4·I·\left(\frac{\sqrt{3}}{4}{a}^2\right)=\sqrt{3}I{a}^2$

Step 3: Calculate the magnetic dipole moment for the square coil
For a square of side $a$:
The perimeter of a single turn is $4a$.
The number of turns $N_2$ that can be made from the wire of length $12a$ is:
$N_2=\frac{12a}{4a}=3$

The area $A_2$ of a square of side $a$ is:
$A_2=a^2$

Using the formula for magnetic moment, we get:
$M_2=N_2·I·A_2=3·I·a^2=3I{a}^2$

Conclusion
The magnetic dipole moments of the coil in each case respectively are:
$\sqrt{3}I{a}^2$ and $3I{a}^2$