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(A) Given: Side of the triangle,$l = 4.5 	imes 10^{-2} \,m$,Current,$I = 1 \,A$.The magnetic field $B$ due to a finite wire at a perpendicular distan

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$4	imes10^{-5}\,Wb/m^2$

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(A) Given: Side of the triangle,$l = 4.5 	imes 10^{-2} \,m$,Current,$I = 1 \,A$.The magnetic field $B$ due to a finite wire at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{4\pi d} (\sin 	heta_1 + \sin 	heta_2)$.For an equilateral triangle,the distance $d$ from the centre to any side is $d = \frac{l}{2\sqrt{3}}$.Substituting $l = 4.5 	imes 10^{-2} \,m$,we get $d = \frac{4.5 	imes 10^{-2}}{2\sqrt{3}} \,m$.For each side,the angles subtended at the centre are $	heta_1 = 	heta_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 I}{4\pi d} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 I}{4\pi d} (2 \sin 60^{\circ}) = \frac{\mu_0 I}{4\pi d} (2 	imes \frac{\sqrt{3}}{2}) = \frac{\mu_0 I \sqrt{3}}{4\pi d}$.Substituting $d = \frac{l}{2\sqrt{3}}$,we get $B_1 = \frac{\mu_0 I \sqrt{3}}{4\pi (l / 2\sqrt{3})} = \frac{\mu_0 I (3)}{2\pi l} = \frac{3 \mu_0 I}{2\pi l}$.The total magnetic field at the centre due to all three sides is $B_{net} = 3 	imes B_1 = 3 	imes \frac{3 \mu_0 I}{2\pi l} = \frac{9 \mu_0 I}{2\pi l}$.Using $\mu_0 = 4\pi 	imes 10^{-7} \,T\cdot m/A$,$I = 1 \,A$,and $l = 4.5 	imes 10^{-2} \,m$:$B_{net} = \frac{9 	imes (4\pi 	imes 10^{-7}) 	imes 1}{2\pi 	imes 4.5 	imes 10^{-2}} = \frac{18 	imes 10^{-7}}{4.5 	imes 10^{-2}} = 4 	imes 10^{-5} \,T$ (or $Wb/m^2$).

$A$ current of $1\,A$ is flowing through the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$. The magnetic field at the centre of the triangle will be

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