Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

$3 I a^2$ and $4 I a^2$

$4 I a^2$ and $3 I a^2$

$\sqrt{3} I a^2$ and $3 I a^2$

$3 I a^2$ and $I a^2$

Step-by-Step Solution

Identify Formula: The magnetic dipole moment ( $M$ ) of a current-carrying coil is given by $M = N I A$ , where $N$ is the number of turns, $I$ is the current, and $A$ is the area of the loop.
Case (i) Equilateral Triangle: Total length of wire $L = 12a$ . Perimeter of one turn (equilateral triangle of side $a$ ) = $3a$ . Number of turns $N_1 = \frac{L}{\text{Perimeter}} = \frac{12a}{3a} = 4$ . Area of equilateral triangle $A_1 = \frac{\sqrt{3}}{4} a^2$ .

Magnetic Moment $M_1 = N_1 I A_1 = 4 \times I \times \frac{\sqrt{3}}{4} a^2 = \sqrt{3} I a^2$ .

Case (ii) Square: Total length of wire $L = 12a$ . Perimeter of one turn (square of side $a$ ) = $4a$ . Number of turns $N_2 = \frac{12a}{4a} = 3$ . Area of square $A_2 = a^2$ .

Magnetic Moment $M_2 = N_2 I A_2 = 3 \times I \times a^2 = 3 I a^2$ .

Conclusion: The magnetic moments are $\sqrt{3} I a^2$ and $3 I a^2$ respectively.

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