Question

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:

• A. 4Ia² and 3Ia²
• B. √3Ia² and 3Ia²
• C. 3Ia² and Ia²
• D. 3Ia² and 4Ia²

Answer 1

The Correct Option is B

To find the magnetic dipole moments of the coils, we need to understand the concept of the magnetic dipole moment for a current-carrying loop. The magnetic dipole moment (\(m\)) is given by the product of current (\(I\)) and the area (\(A\)) of the loop:

\(m = I \times A\)

Let’s break down the problem for both cases: 

1. Equilateral Triangle:
   • The length of the wire is \(12a\), and it is formed into an equilateral triangle with side length \(a\).
   • The perimeter of the triangle is \(3a\). Thus, the wire can form exactly 360° of the triangle.
   • The area (\(A\)) of an equilateral triangle is given by: \(A = \frac{\sqrt{3}}{4} a^2\)
   • Using the formula for magnetic dipole moment: \(m = I \times \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} Ia^2\)
   • Since the question involves two full loops for creating a coil, the magnetic dipole moment becomes: \(2 \times \frac{\sqrt{3}}{4} Ia^2 = \frac{\sqrt{3}}{2} Ia^2\)
2. Square:
   • Using the same wire of \(12a\), it is now formed into a square with side \(a\).
   • The perimeter of the square is \(4a\).
   • The area (\(A\)) of the square is: \(A = a^2\)
   • Using the formula for magnetic dipole moment: \(m = I \times a^2\)
   • Since the wire completes \(12a/4a = 3\) loops around the square, we have: \(3 \times Ia^2 = 3Ia^2\)

Therefore, the magnetic dipole moments of the coil when wound in the shape of an equilateral triangle and a square are \(\sqrt{3}Ia^2\) and \(3Ia^2\) respectively.

Hence, the correct answer is: √3Ia² and 3Ia²