Question

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 V\(_0\). The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1 : \(\sqrt{y}\) where y is _________.

Hint:

For a given length of wire turned into a multi-turn coil, the magnetic moment is \(M = N I A = (\frac{L}{P}) I A\), where P is the perimeter. This shows that for a fixed side length, the number of turns plays a crucial role. Always calculate N first.

Answer 1

Correct Answer: 3

Step 1: Understanding the Question:
A wire of fixed length is first used to make a triangular coil and then a square coil, both with side length 'a'. We need to find the ratio of their magnetic dipole moments.
Step 2: Key Formula or Approach:
The magnetic dipole moment (\(M\)) of a current loop is given by \(M = NIA\), where: - \(N\) is the number of turns in the coil. - \(I\) is the current flowing through the coil. - \(A\) is the area of the loop. The number of turns \(N\) is the total length of the wire \(L\) divided by the perimeter of one loop. The current \(I\) is the same for both cases since the wire (with resistance R) is connected to the same voltage source \(V_0\). Thus, \(I = V_0/R\).
Step 3: Detailed Explanation:
Total length of the wire, \(L = 24a\).
Case 1: Equilateral Triangle Coil
Side of the triangle = \(a\). Perimeter of one turn = \(3a\). Number of turns, \(N_t = \frac{L}{\text{Perimeter}} = \frac{24a}{3a} = 8\). Area of an equilateral triangle, \(A_t = \frac{\sqrt{3}}{4}a^2\). Magnetic moment of the triangular coil: \[ M_{triangle} = N_t I A_t = 8 \times I \times \left(\frac{\sqrt{3}}{4}a^2\right) = 2\sqrt{3} I a^2 \] Case 2: Square Coil
Side of the square = \(a\). Perimeter of one turn = \(4a\). Number of turns, \(N_s = \frac{L}{\text{Perimeter}} = \frac{24a}{4a} = 6\). Area of the square, \(A_s = a^2\). Magnetic moment of the square coil: \[ M_{square} = N_s I A_s = 6 \times I \times (a^2) = 6 I a^2 \] Ratio of Magnetic Moments
We need to find the ratio \(\frac{M_{triangle}}{M_{square}}\). \[ \frac{M_{triangle}}{M_{square}} = \frac{2\sqrt{3} I a^2}{6 I a^2} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \] The question states this ratio is \(1 : \sqrt{y}\), which means \(\frac{1}{\sqrt{y}}\). \[ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{y}} \] By comparison, we get \(y=3\).
Step 4: Final Answer:
The value of y is 3.