Two identical wires A and B each of length $‘ l ’$ , carry the same current $‘ I ’$ . Wire A is bent into a circle of radius R and wire B is bent to form a square of side $' a'$ . If $B_{A}$ and $B_{B}$ are the values of magnetic field at the centres of the circle and square respectively, then the ratio $\frac{B_{A}}{B_{B}} is \frac{π^{2}}{x \sqrt{2}} .$ Then the value of x is........

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Detailed Solution

wire A bent into a circle of radius R,

$l = 2 π R \Rightarrow R = \frac{l}{2 π}$

$B_{A} = \frac{μ_{0} I}{2 R} = \frac{μ_{0} I}{2 \times (\frac{l}{2 π})} = \frac{μ_{0} π I}{l}$

Wire B is bent into a square of side a,

$l = 4 a \Rightarrow a = \frac{l}{4}$

$B_{B} = 4 \times [\frac{μ_{0} I}{4 π (a / 2)} (\sin 45^{\circ} + \sin 45^{\circ})]$

$B_{B} = \frac{16 μ_{0} I}{\sqrt{2} π l},$

$\frac{B_{A}}{B_{B}} = \frac{π^{2}}{8 \sqrt{2}}$

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