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 :

Correct Option: Option 3: √ 3 Ia² and 3 Ia²

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:

$M=N·I·A$

The total length of the wire is $L=12a$.

Case (i): Equilateral triangle of side ‘a’

Perimeter of one triangular turn = $3a$

Number of turns, $N_1=\frac{12a}{3a}=4$

Area of an equilateral triangle of side $a$, $A_1=\frac{\sqrt{3}}{4}{a}^2$

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’

Perimeter of one square turn = $4a$

Number of turns, $N_2=\frac{12a}{4a}=3$

Area of a square of side $a$, $A_2=a^2$

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 √ 3 Ia² and 3 Ia².