A uniform conducting wire of length '12a' and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then a square of side 'a'. The ratio of magnetic dipole moments of triangular to square coil is (a) 1: √(3) (b) √(3): 1 (c) 1: 3 (d) 3: 1

Answer: (a). Same wire, same resistance, so the same current I flows in both coils. M = NIA. Triangle: perimeter 3a, so N = 12a/3a = 4; Mt = 4I·√(3)/4a² = √(3)Ia². Square: perimeter 4a, so N = 3; Ms = 3Ia². (Mt)/(Ms) = √(3)/3 = 1/√(3), so the answer is (a) 1: √(3).

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Current Loop as a Magnetic Dipole

A uniform conducting wire of length $\text{'}12a\text{'}$ and resistance $R$ is wound up as a current carrying coil in the shape of an equilateral triangle of side $\text{'}a\text{'}$ and then a square of side $\text{'}a\text{'}$ . The ratio of magnetic dipole moments of triangular to square coil is

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Answer: (a)

Same wire, same resistance, so the same current $I$ flows in both coils. $M = NIA$ .
Triangle: perimeter $3a$ , so $N = \frac{12a}{3a} = 4$ ; $M_t = 4I\cdot\frac{\sqrt{3}}{4}a^2 = \sqrt{3}Ia^2$ .
Square: perimeter $4a$ , so $N = 3$ ; $M_s = 3Ia^2$ .
$\frac{M_t}{M_s} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$ , so the answer is (a) $1 : \sqrt{3}$ .

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