A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then in the form of a square of side 'a'. The coil is connected to a voltage source V₀. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1 : √y where y is ................. .

Solution

Physics sir · 3 steps

Same wire means same current; the shape only changes the number of turns and the area of each turn.

1. Same current. Both coils use the same wire with resistance R, across the same V₀. So the current I = V₀/R is the same in both cases.
2. Count the turns. One triangular turn uses 3a of wire, so the triangle coil has N₁ = 24a/3a = 8 turns. One square turn uses 4a, so the square coil has N₂ = 24a/4a = 6 turns.

   M₁/M₂ = (N₁ I A₁)/(N₂ I A₂) = (8 × √3/4a²)/(6 × a²) = 2√3/6 = 1/√3

3. So. The ratio is 1 : √3, so y = 3.

The common slip: 5.33. Comparing only the areas of one turn and forgetting that the coils have 8 and 6 turns gives √3/4 : 1, which is y = 16/3 ≈ 5.33, not 3.

The idea to keep. For a coil, M = NIA. When a fixed length of wire is rewound into a new shape, work out the number of turns from the perimeter first, then multiply by the area of one turn.

Answer from the NTA final answer key. Solution by PrepWiser.