Question

A wire of length $12L$ is converted into a equilateral triangle and a square. Find the ratio of magnetic field at the centroid of triangle and centre of square if both are carrying current $I$.

• A. $\frac{27}{4\sqrt{2}}$
• B. $\frac{27}{16}$
• C. $\frac{5}{16}$
• D. $\frac{9}{10\sqrt{2}}$

Answer 1

The correct option is A $\frac{27}{4\sqrt{2}}$

The magnetic field due to the triangle at its centroid is,


Magnetic field due to side $\text{A}B$ is ,

$\because$ Point $\text{O}$ is on the perpendicular bisector of side $\text{A}B$, field at $\text{O}$ will be,

$B_{AB}=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

$B_{AB}=\frac{{\mu }_{0}I}{4\pi d}\left(2\sin \theta \right)[\because \sin {\theta }_1=\sin {\theta }_2)]$

From the $△AOD$ ,

$\tan {30}^{\circ }=\frac{d}{2L}\Rightarrow d=\frac{2L}{\sqrt{3}}$

$B_{AB}=\frac{{\mu }_{0}I}{4\pi \left(\frac{2L}{\sqrt{3}}\right)}\left(2\sin {60}^{\circ }\right)$

$=\frac{{\mu }_{0}I}{4\pi \left(\frac{2L}{\sqrt{3}}\right)}(2\times \frac{\sqrt{3}}{2})$

$=\frac{6{\mu }_{0}I}{4\pi L}$

Field at the centroid of the triangle, due to all three sides is,

$B_{\text{net}}=3B_{AB}=\frac{18{\mu }_{0}I}{4\pi L}$

$B_T=\frac{18{\mu }_0}{4\pi }\frac{I}{L}...........\left(1\right)$

Magnetic field due to the square at the centre is,


Magnetic field due to side $\text{A}D$ is ,

$\because$ Point $\text{O}$ is on the perpendicular bisector of side $\text{A}D$, field at $\text{O}$ will be,

$B_{AD}=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

$B_{AD}=\frac{{\mu }_{0}I}{4\pi d}\left(2\sin \theta \right)[\because \sin {\theta }_1=\sin {\theta }_2)]$

$B_{AD}=\frac{{\mu }_{0}I}{4\pi \left(\frac{3L}{2}\right)}\left(2\sin {45}^{\circ }\right)$

$=\frac{{\mu }_{0}I}{4\pi \left(\frac{3L}{2}\right)}(2\times \frac{1}{\sqrt{2}})$

$=\frac{{\mu }_{0}I}{4\pi L}\left(\frac{2\sqrt{2}}{3}\right)$

Field at the centre of the square, due to all four sides is,

$B_{\text{net}}=4B_{AD}=\frac{{\mu }_{0}I}{4\pi L}\left(\frac{8\sqrt{2}}{3}\right)...........\left(2\right)$

$B_T=\frac{18{\mu }_0}{4\pi }\frac{I}{L}$

On dividing (1)$÷$(2) we get,

$\frac{B_T}{B_S}=\frac{\frac{{\mu }_{0}I}{4\pi L}\left(18\right)}{\frac{{\mu }_{0}I}{4\pi L}\left(\frac{8\sqrt{2}}{3}\right)}$

$=\frac{3\left(18\right)}{8\sqrt{2}}=\frac{27}{4\sqrt{2}}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.