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(D) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4\pi r} (\sin 	heta_1

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$18$

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(D) The magnetic field $B$ due to a straight wire of length $L$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i}{4\pi r} (\sin 	heta_1 + \sin 	heta_2)$.For an equilateral triangle of side $a = 1\,m$,the distance $r$ from the center to the side is $r = \frac{a}{2	an(60^\circ)} = \frac{a}{2\sqrt{3}}$.For each side,$	heta_1 = 	heta_2 = 60^\circ$,so $\sin 	heta_1 + \sin 	heta_2 = 2 \sin(60^\circ) = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4\pi (a/2\sqrt{3})} 	imes \sqrt{3} = \frac{\mu_0 i \sqrt{3}}{2\pi a} 	imes \sqrt{3} = \frac{3\mu_0 i}{2\pi a}$.Since there are $3$ sides,the total magnetic field at the center is $B = 3 	imes B_1 = 3 	imes \frac{3\mu_0 i}{2\pi a} = \frac{9\mu_0 i}{2\pi a}$.Substituting the values: $B = \frac{9 	imes (4\pi 	imes 10^{-7}) 	imes 10}{2\pi 	imes 1} = 18 	imes 10^{-6}\,T = 18\,\mu T$.

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side $1\,m$ which is carrying a current of $10\,A$ is:......$\mu T$ [Take $\mu _0 = 4\pi \times 10^{-7}\,NA^{-2}$]

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