Understanding Magnetic Dipole Moment Calculation for Coils

This problem involves calculating the magnetic dipole moment for a current-carrying wire shaped into two different configurations: a square coil and a regular hexagonal coil. The magnetic dipole moment ($ \vec{m} $) of a current loop is a measure of its magnetic strength and orientation. Its magnitude is given by the formula $ m = NIA $, where $ N $ is the number of turns in the coil, $ I $ is the current flowing through the wire, and $ A $ is the area enclosed by one turn of the coil.

We are given a uniform conducting wire of total length $ L = 12a $ and resistance $ R $, carrying a current $ I $. We need to find the magnetic dipole moment for two cases:

• Case (i): A square coil where each side has length $ s_1 = \frac{12a}{16} $.
• Case (ii): A regular hexagonal coil where each side has length $ s_2 = \frac{12a}{24} $.

The resistance $ R $ is provided but not needed for calculating the magnetic dipole moment. The key is to determine the number of turns ($ N $) and the area ($ A $) for each shape using the total wire length $ 12a $.

Calculating Magnetic Dipole Moment for the Square Coil

First, let's analyze the square coil.

• Side length ($ s_1 $): Given as $ s_1 = \frac{12a}{16} = \frac{3a}{4} $.
• Perimeter ($ P_1 $): The perimeter of one square turn is $ P_1 = 4 \times s_1 = 4 \times \frac{3a}{4} = 3a $.
• Number of Turns ($ N_1 $): The total length of the wire is $ L = 12a $. The length required for one turn is $ P_1 = 3a $. Therefore, the number of turns is $ N_1 = \frac{L}{P_1} = \frac{12a}{3a} = 4 $.
• Area ($ A_1 $): The area of one square turn is $ A_1 = s_1^2 = (\frac{3a}{4})^2 = \frac{9a^2}{16} $.
• Magnetic Dipole Moment ($ m_1 $): Using the formula $ m = NIA $, the magnetic dipole moment for the square coil is $ m_1 = N_1 I A_1 = 4 \times I \times \frac{9a^2}{16} $.

Simplifying the expression for $ m_1 $: $ m_1 = \frac{4 \times 9}{16} I a^2 = \frac{36}{16} I a^2 = \frac{9}{4} I a^2 $.

Calculating Magnetic Dipole Moment for the Hexagonal Coil

Next, let's analyze the regular hexagonal coil.

• Side length ($ s_2 $): Given as $ s_2 = \frac{12a}{24} = \frac{a}{2} $.
• Perimeter ($ P_2 $): A regular hexagon has 6 sides. The perimeter of one hexagonal turn is $ P_2 = 6 \times s_2 = 6 \times \frac{a}{2} = 3a $.
• Number of Turns ($ N_2 $): The total length of the wire is $ L = 12a $. The length required for one turn is $ P_2 = 3a $. Therefore, the number of turns is $ N_2 = \frac{L}{P_2} = \frac{12a}{3a} = 4 $.
• Area ($ A_2 $): The area of a regular hexagon with side length $ s $ is given by $ A = \frac{3\sqrt{3}}{2} s^2 $. For this coil, the area of one hexagonal turn is $ A_2 = \frac{3\sqrt{3}}{2} s_2^2 = \frac{3\sqrt{3}}{2} (\frac{a}{2})^2 $.

Calculating the area $ A_2 $: $ A_2 = \frac{3\sqrt{3}}{2} \times \frac{a^2}{4} = \frac{3\sqrt{3}}{8} a^2 $.

Magnetic Dipole Moment ($ m_2 $): Using the formula $ m = NIA $, the magnetic dipole moment for the hexagonal coil is $ m_2 = N_2 I A_2 = 4 \times I \times \frac{3\sqrt{3}}{8} a^2 $.

Simplifying the expression for $ m_2 $: $ m_2 = \frac{4 \times 3\sqrt{3}}{8} I a^2 = \frac{12\sqrt{3}}{8} I a^2 = \frac{3\sqrt{3}}{2} I a^2 $.

Summary of Magnetic Dipole Moments

The calculated magnetic dipole moments for the square coil and the hexagonal coil are:

• Square coil: $ m_1 = \frac{9}{4} I a^2 $
• Hexagonal coil: $ m_2 = \frac{3\sqrt{3}}{2} I a^2 $

Comparing these results with the given options, the correct pair is $ \frac{9}{4} I a^2 $ and $ \frac{3\sqrt{3}}{2} I a^2 $.