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 √3Ia² and 3Ia².

Step-by-Step Explanation:

The magnetic dipole moment ($M$) of a current-carrying coil with $N$ turns, carrying current $I$, and enclosing an area $A$ is given by the formula:
$M=N·I·A$

Given details:
Total length of the wire = $12a$
Let the current carrying through the coil in each case be $I$.

Case (i): Equilateral triangle of side 'a'
1. The perimeter of a single equilateral triangle of side $a$ is:
$P_1=3a$
2. The number of turns ($N_1$) made from the wire of total length $12a$ is:
$N_1=\frac{12a}{3a}=4$
3. The area ($A_1$) of an equilateral triangle of side $a$ is:
$A_1=\frac{\sqrt{3}}{4}{a}^2$
4. Therefore, the magnetic dipole moment ($M_1$) for the triangular coil is:
$M_1=N_1·I·A_1=4·I·\left(\frac{\sqrt{3}}{4}{a}^2\right)=\sqrt{3}I{a}^2$

Case (ii): Square of side 'a'
1. The perimeter of a single square of side $a$ is:
$P_2=4a$
2. The number of turns ($N_2$) made from the wire of total length $12a$ is:
$N_2=\frac{12a}{4a}=3$
3. The area ($A_2$) of a square of side $a$ is:
$A_2=a^2$
4. Therefore, the magnetic dipole moment ($M_2$) for the square coil is:
$M_2=N_2·I·A_2=3·I·a^2=3I{a}^2$

Thus, the magnetic dipole moments of the coil in each case respectively are √3Ia² and 3Ia².