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(A) The magnetic field at the centre of a square loop of side $a$ carrying current $i$ is given by the formula:$B = 4 	imes \left( \frac{\mu_0 i}{4 \

Vedclass · Accepted Answer

$12 \sqrt{2} 	imes 10^{-6} \, T$

Vedclass · Answer

(A) The magnetic field at the centre of a square loop of side $a$ carrying current $i$ is given by the formula:$B = 4 	imes \left( \frac{\mu_0 i}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2) ight)$For a square loop,the distance from the centre to the side is $d = a/2 = 0.1 \, m$.The angles at the centre subtended by the half-side are $	heta_1 = 	heta_2 = 45^\circ$.Thus,$B = 4 	imes \frac{\mu_0 i}{4 \pi (a/2)} (\sin 45^\circ + \sin 45^\circ)$$B = \frac{2 \mu_0 i}{\pi a} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{2 \mu_0 i}{\pi a} 	imes \frac{2}{\sqrt{2}} = \frac{2 \sqrt{2} \mu_0 i}{\pi a}$Given $i = 3 \, A$,$a = 0.2 \, m$,and $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$:$B = \frac{2 \sqrt{2} 	imes (4 \pi 	imes 10^{-7}) 	imes 3}{\pi 	imes 0.2}$$B = \frac{2 \sqrt{2} 	imes 4 	imes 10^{-7} 	imes 3}{0.2} = \frac{24 \sqrt{2} 	imes 10^{-7}}{0.2} = 120 \sqrt{2} 	imes 10^{-7} \, T = 12 \sqrt{2} 	imes 10^{-6} \, T$.

Find the magnetic field at the centre $P$ of a square loop of side length $20 \, cm$ carrying a current of $3 \, A$.

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