Q149 / 180 · COMEDK 2022 · Physics
A regular hexagon of side $m$ which is a wire of length $24 \mathrm{~m}$ is coiled on that hexagon. If current in hexagon is $I$, then the magnetic moment,
- Correct answerA$6 \sqrt{3} \operatorname{Im}^2$
- B$3 \sqrt{3} \mathrm{~lm}^2$
- C$\dfrac{3 \sqrt{3}}{2} \operatorname{lm}^2 \quad$
- D$6 \operatorname{lm}^2$
Solution
The magnetic moment $\vec{\mu}$ of a planar current loop is given by $\mu = N I A$, where $N$ is the number of turns, $I$ is the current, and $A$ is the area of the loop.
The total length of the wire is $L = 24m$. The perimeter of a regular hexagon with side length $m$ is $P = 6m$.
The number of turns $N$ is the total length of the wire divided by the perimeter of one hexagon:
$N = \dfrac{L}{P} = \dfrac{24m}{6m} = 4$.
The area $A$ of a regular hexagon with side length $m$ is given by:
$A = \dfrac{3\sqrt{3}}{2} m^2$.
Substituting these values into the formula for magnetic moment:
$\mu = N I A = 4 \times I \times \left( \dfrac{3\sqrt{3}}{2} m^2 \right)$
$\mu = 2 \times 3\sqrt{3} I m^2 = 6\sqrt{3} I m^2$.
Answer: $6 \sqrt{3} \operatorname{Im}^2$
The total length of the wire is $L = 24m$. The perimeter of a regular hexagon with side length $m$ is $P = 6m$.
The number of turns $N$ is the total length of the wire divided by the perimeter of one hexagon:
$N = \dfrac{L}{P} = \dfrac{24m}{6m} = 4$.
The area $A$ of a regular hexagon with side length $m$ is given by:
$A = \dfrac{3\sqrt{3}}{2} m^2$.
Substituting these values into the formula for magnetic moment:
$\mu = N I A = 4 \times I \times \left( \dfrac{3\sqrt{3}}{2} m^2 \right)$
$\mu = 2 \times 3\sqrt{3} I m^2 = 6\sqrt{3} I m^2$.
Answer: $6 \sqrt{3} \operatorname{Im}^2$
Hence, the correct answer is option: 'A'.