Question

A uniform conducting wire of length $18a$ and resistance $R$ is wound up as current carrying coil in the shape of a regular 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

Answer: (B)

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