Question

Two identical wires $\text{A}$ and $\text{B}$ each of length $L$, carry the same current $I$. Wire $\text{A}$ is bent into a circle of radius $R$ and wire $\text{B}$, is bent to form a square of side $a$. If $B_{\text{A}}$ and $B_{\text{B}}$ are the values of magnetic fields at the center of circle and square then the ratio of $\frac{B_{\text{A}}}{B_{\text{B}}}$ is

• A. $\frac{{\pi }^2}{8}$
• B. $\frac{{\pi }^2}{16\sqrt{2}}$
• C. $\frac{{\pi }^2}{16}$
• D. $\frac{{\pi }^2}{8\sqrt{2}}$

Answer 1

The correct option is D $\frac{{\pi }^2}{8\sqrt{2}}$


As the wire $\text{A}$ is bent into a circle of radius $R$ so,

$2\pi R=L\Rightarrow R=\frac{L}{2\pi }$

So the magnetic field at the center of ring is given by,

$B_{\text{A}}=\frac{{\mu }_{0}I}{2R}$

$B_A=\frac{{\mu }_{0}I}{2\times \frac{L}{2\pi }}=\frac{2\pi {\mu }_{0}I}{2L}........\left(1\right)$

Again another wire of length $L$ is made to form an square of side $a$. So,

$4a=L,\text{so}a=\frac{L}{4}$

Magnetic field due to wire segment $\text{PQ}$ at point $O$ will be

$B_{PQ}=\frac{{\mu }_{0}I}{4\pi \left(OS\right)}(\sin {\theta }_1+\sin {\theta }_2)$

Here, ${\theta }_1={\theta }_2={45}^{\circ };OS=a/2$

$B_{PQ}=\frac{{\mu }_{0}I}{4\pi \left(a/2\right)}(\sin {45}^{\circ }+\sin {45}^{\circ })$

$B_{PQ}=\frac{{\mu }_0}{4\pi }\frac{2I}{a}\times \sqrt{2}$

So, net magnetic field due to square will be

$B_B=4B_{PQ}=4\times \frac{{\mu }_0}{4\pi }\frac{2I}{a}\times \sqrt{2}=8\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{a}$

$B_B=8\sqrt{2}\frac{{\mu }_0}{4\pi }\frac{I}{L/4}[\because a=\frac{L}{4}]$

$B_B=8\sqrt{2}\frac{{\mu }_0}{\pi }\frac{I}{L}........\left(2\right)$

From equations $\left(1\right)\&\left(2\right)$

$\frac{B_{\text{A}}}{B_{\text{B}}}=\frac{2\pi {\mu }_{0}I}{2L}/8\sqrt{2}\frac{{\mu }_0}{\pi }\frac{I}{L}$

$\therefore \frac{B_{\text{A}}}{B_{\text{B}}}=\frac{{\pi }^2}{8\sqrt{2}}$

Hence, option $\left(d\right)$ is correct.