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A square frame of side 1 m carries a current I, produces a magnetic field B at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is B'. The ratio B l B' is

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a

$\frac{8}{π^{2}}$

b

$\frac{8 \sqrt{2}}{π^{2}}$

c

$\frac{16}{π^{2}}$

d

$\frac{16}{\sqrt{2} π^{2}}$

answer is B.

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Detailed Solution

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

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$l = 4 a \Rightarrow a = \frac{l}{4} B' = \frac{μ_{0} i}{2 R} = \frac{μ_{0} i}{2 \cdot \frac{l}{2 π}} = \frac{μ_{0} π i}{l} B = 4 [\frac{μ_{0}}{4 π} \cdot \frac{i}{a / 2} (\sin 45^{\circ} + \sin 45^{\circ})] = \frac{2 \sqrt{2} μ_{0} i}{π a} = \frac{8 \sqrt{2} μ_{0} i}{π l} \frac{B}{B^{'}} = \frac{8 \sqrt{2}}{π^{2}}$

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