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 $I=\frac{V_{0}}{R})$