Question

A uniform conducting wire of length 18a and resistance R is wound up as current-carrying coil in the shape of a regular a hexagon of sides a. If the coil is connected to a voltage source $V_0$, then the magnetic moment of coil is:

• A. $6\sqrt{3}\frac{V_0}{R}{a}^{2}A{m}^2$
• B. $9\frac{\sqrt{3}}{2}\frac{V_0{a}^2}{R}A{m}^2$
• C. $\frac{7\sqrt{3}}{2}\frac{V_0{a}^2}{R}A{m}^2$
• D. $\frac{11\sqrt{3}}{2}\frac{V_0{a}^2}{R}A{m}^2$

Answer 1

The correct option is B $9\frac{\sqrt{3}}{2}\frac{V_0{a}^2}{R}A{m}^2$

In the regular hexagon, if each arm length is a, then the number of turns in the given shape, n = $\frac{18a}{6a}$= 3
Now the area of given shape is A = $\frac{3\sqrt{3}{a}^2}{2}$
Now magnetic moment M = $nIA$

$M=3$ x $I$ x $\frac{3\sqrt{3}{a}^2}{2}=9\frac{\sqrt{3}}{2}\frac{V_0{a}^2}{R}A{m}^2$ $(whereI=\frac{V_0}{R})$