Question

The self inductance of an equilateral triangular coil having $N$ number of turns as shown in the figure

• A. $\frac{9\sqrt{3}}{8}\frac{{\mu }_0{N}^{3}a}{\pi }$
• B. $\frac{3\sqrt{3}}{4}\frac{{\mu }_0{N}^{3}a}{\pi }$
• C. $\frac{9\sqrt{3}}{4}\frac{{\mu }_0{N}^{3}a}{\pi }$
• D. $\frac{3\sqrt{3}}{2}\frac{{\mu }_0{N}^{3}a}{\pi }$

Answer 1

The correct option is A $\frac{9\sqrt{3}}{8}\frac{{\mu }_0{N}^{3}a}{\pi }$

Self inductance for a loop having $N$ turns is given by,

$L=\frac{N{\varphi }_B}{i}$

Where, ${\varphi }_B$ is the magnetic flux and $i$ is the current flowing in the loop.


From figure, the distance $OA$ is,

$d=\frac{a}{2}\tan {30}^0=\frac{a}{2\sqrt{3}}$

Magnetic field due to $DB$ at $O$

$B_1=\frac{{\mu }_{0}i}{4\pi d}[\cos {30}^{\circ }+\cos {30}^{\circ }]$

$=\frac{{\mu }_{0}i\times 2\sqrt{3}\times \sqrt{3}}{4\pi \times a}=\frac{3{\mu }_{0}i}{2\pi a}$

$\therefore \text{Net magnetic field},B=3B_1$

$B=\frac{3\times 3\times {\mu }_{0}i\times N}{2\pi a}=\frac{9{\mu }_{0}iN}{2\pi a}$

$\therefore$ Magnetic flux is given by,

${\varphi }_B=BA$

$=\frac{9{\mu }_{0}iN}{2\pi a}\times \frac{1}{2}\times a\times a\sin {60}^{\circ }$

$=\frac{9\sqrt{3}{\mu }_{0}iaN}{8\pi }$

$\therefore$ Self inductance of coil is,

$L=\frac{N{\varphi }_B}{i}$

$\Rightarrow L=\frac{N}{i}\times \frac{9\sqrt{3}}{8}\frac{{\mu }_{0}Nia}{\pi }$

$\therefore L=\frac{9\sqrt{3}}{8}\frac{{\mu }_0{N}^{2}a}{\pi }$

Hence, option $\left(\text{A}\right)$ is correct.