A current of 1 ,A is flowing along the sides | vedclass

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(A) For a straight wire of length $a$ carrying current $i$, the magnetic field at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r

Vedclass · Accepted Answer

$4 	imes 10^{-5} \,T$

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(A) For a straight wire of length $a$ carrying current $i$, the magnetic field at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4 \pi r} (\sin \phi_1 + \sin \phi_2)$.For an equilateral triangle of side $a$, the distance $r$ from the centroid to any side is $r = \frac{a}{2 \sqrt{3}}$.The angles subtended by the corners at the centroid are $\phi_1 = \phi_2 = 60^{\circ}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi (a / 2 \sqrt{3})} (\sin 60^{\circ} + \sin 60^{\circ}) = \frac{\mu_0 i}{4 \pi (a / 2 \sqrt{3})} (\sqrt{3}) = \frac{\mu_0 i \sqrt{3}}{4 \pi (a / 2 \sqrt{3})} = \frac{3 \mu_0 i}{2 \pi a}$.The total magnetic field at the centroid due to all three sides is $B = 3 B_1 = \frac{9 \mu_0 i}{2 \pi a}$.Given $i = 1 \,A$, $a = 4.5 	imes 10^{-2} \,m$, and $\frac{\mu_0}{4 \pi} = 10^{-7} \,T \cdot m/A$, we have:$B = 3 	imes \frac{10^{-7} 	imes 1 	imes (\sqrt{3} + \sqrt{3})}{a / 2 \sqrt{3}} = 3 	imes \frac{10^{-7} 	imes 2 \sqrt{3}}{4.5 	imes 10^{-2} / 2 \sqrt{3}} = 3 	imes \frac{10^{-7} 	imes 2 \sqrt{3} 	imes 2 \sqrt{3}}{4.5 	imes 10^{-2}} = 3 	imes \frac{10^{-7} 	imes 12}{4.5 	imes 10^{-2}} = \frac{36 	imes 10^{-5}}{4.5} = 8 	imes 10^{-5} \,T$.Wait, re-calculating: $B = 3 	imes \frac{\mu_0 i}{4 \pi r} (2 \sin 60^{\circ}) = 3 	imes \frac{10^{-7} 	imes 1}{4.5 	imes 10^{-2} / 2 \sqrt{3}} 	imes \sqrt{3} = 3 	imes \frac{10^{-7} 	imes 2 \sqrt{3} 	imes \sqrt{3}}{4.5 	imes 10^{-2}} = 3 	imes \frac{10^{-7} 	imes 6}{4.5 	imes 10^{-2}} = \frac{18 	imes 10^{-5}}{4.5} = 4 	imes 10^{-5} \,T$.

$A$ current of $1 \,A$ is flowing along the sides of an equilateral triangle of side $a = 4.5 \times 10^{-2} \,m$. The magnetic field at the centroid of the triangle is $(\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A)$.

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