JEE MainPhysicsEasy / Moderate/ ToughMagnetic effect of current

Two identical wires A and B, each of length $\mathcal{l}$ , carry
the same current I. Wire A is bent into acircle of radius R and wire B
is bent form a squareof side a. If B_Aand B_B are the values of
magnetic field at the centres of the circle and square respectively,
then the ratio B_A/ B_B is $\pi^{2}:n\sqrt{2}.\$ What is the value of
n.

Show answer and solution

B at centre of a circle

= \frac{\mu_{0}I}{2R}

B at centre of a
square

= 4 \times \frac{\mu I}{4\pi \cdot \frac{\mathcal{l}}{2}}\left( sin45^{\circ} + sin45^{\circ} \right)

= 4\sqrt{2}\frac{\mu_{0}I}{2\pi\mathcal{l}}

Now,

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

and

\mathcal{l =}\frac{L}{4}(\ as\ L = 2\pi R = 4\mathcal{l)}

Where, L= length of wire

\therefore{\ \ \ B}_{A} = \frac{\mu_{0}L}{2 \cdot \frac{L}{2\pi}} = \frac{\pi\mu_{0}I}{L} = \pi\left( \frac{\mu_{0}I}{L} \right)

B_{B} = 4\sqrt{2}\frac{\mu_{0}I}{2\pi\left( \frac{L}{4} \right)} = \frac{8\sqrt{2}\mu_{0}I}{\pi L} = \frac{8\sqrt{2}}{\pi}\left( \frac{\mu_{0}I}{L} \right)

\therefore{\ \ \ \ B}_{A}:B_{B} = \pi^{2}:8\sqrt{2}

Source: Grass-G / P-13 Physics paper 4 Nov.docx

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