Ncert Exercises & Solutions

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 sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source $V_{0}$ $V_{0}$ . Find the magnetic moment of the coils in each case.

Hint: The magnetic moment depends on the area of the coil.

We know that the magnetic moment of the coils, M= nlA

Since the same wire is used in three cases with the same potentials, therefore, the same current flows in three cases.

Step 1:

(i) For an equilateral triangle of side a,
n= 4 as the total wire of length 12a

\begin{matrix} Magnetic moment of the coil, M = nIA = 4 I (\frac{\sqrt{3}}{4} a^{2}) ∴ M = {Ia}^{2} \sqrt{3} \end{matrix}

$\begin{array}{l} Magnetic moment of the coil, M = nIA = 4 I (\frac{\sqrt{3}}{4} a^{2}) \\ ∴ M = {Ia}^{2} \sqrt{3} \end{array}$

(ii) Step 2: For a square of sides a,

\begin{matrix} n = 3 as the total wire of length = 12 a Magnetic moment of the coils, M = nIA = 3 I (a^{2}) = 3 {Ia}^{2} \end{matrix}

$\begin{array}{l} n = 3 as the total wire of length = 12 a \\ Magnetic moment of the coils, M = nIA = 3 I (a^{2}) = 3 {Ia}^{2} \end{array}$

(iii) Step 3: For a regular hexagon of sides a,

\begin{matrix} n = 2 as the total wire of length = 12 a Magnetic moment of the coils, M = nIA = 2 I (\frac{6 \sqrt{3}}{4} a^{2}) M = 3 \sqrt{3} a^{2} I \end{matrix}

$\begin{array}{l} n = 2 as the total wire of length = 12 a \\ Magnetic moment of the coils, M = nIA = 2 I (\frac{6 \sqrt{3}}{4} a^{2}) \\ M = 3 \sqrt{3} a^{2} I \end{array}$

M is in a geometric series.

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