Search any question & find its solution
Question:
Answered & Verified by Expert
A regular hexagone of side a. A wire of length 24 a is coiled on that hexagone. If current in hexagone is I, then find the magnetic moment.
Options:
Solution:
1751 Upvotes
Verified Answer
The correct answer is:
$6 \sqrt{3} a^2$
Let number of turns $=\mathrm{n}$
$\begin{aligned} & n \times 6 a=24 a \\ & \therefore \quad n=4 \\ & \end{aligned}$
Magnetic moment, $\mathrm{M}=\mathrm{nIA}=$ 4IA
$\begin{aligned} & \because \text { Area of hexagone }=\frac{1}{2} \mathrm{a}^2 \sin 60^{\circ}+\mathrm{a} \cdot 2 \mathrm{a} \sin 60^{\circ} \\ & \quad+\frac{1}{2} \mathrm{a}^2 \sin 120^{\circ}\end{aligned}$
$=\frac{\sqrt{3}}{4} a^2+\sqrt{3} a^2+\frac{\sqrt{3} a^2}{4}=\frac{6 \sqrt{3} a^2}{4}=\frac{3 \sqrt{3} a^2}{2}$
Hence, magnetic moment $=4 I\left(\frac{3 \sqrt{3} a^2}{2}\right)=6 \sqrt{3} a^2$
$\begin{aligned} & n \times 6 a=24 a \\ & \therefore \quad n=4 \\ & \end{aligned}$
Magnetic moment, $\mathrm{M}=\mathrm{nIA}=$ 4IA
$\begin{aligned} & \because \text { Area of hexagone }=\frac{1}{2} \mathrm{a}^2 \sin 60^{\circ}+\mathrm{a} \cdot 2 \mathrm{a} \sin 60^{\circ} \\ & \quad+\frac{1}{2} \mathrm{a}^2 \sin 120^{\circ}\end{aligned}$
$=\frac{\sqrt{3}}{4} a^2+\sqrt{3} a^2+\frac{\sqrt{3} a^2}{4}=\frac{6 \sqrt{3} a^2}{4}=\frac{3 \sqrt{3} a^2}{2}$
Hence, magnetic moment $=4 I\left(\frac{3 \sqrt{3} a^2}{2}\right)=6 \sqrt{3} a^2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.