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

Vedclass · Accepted Answer

$40$

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(B) The magnetic field due to a straight wire of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 i}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$.For a square loop of side $a = 1 \ m$,the distance from the center to any side is $d = a/2 = 0.5 \ m$.The angles at the center subtended by the corners are $	heta_1 = 	heta_2 = 45^\circ$,so $\sin 45^\circ = 1/\sqrt{2}$.The magnetic field due to one side is $B_1 = \frac{\mu_0 i}{4 \pi (a/2)} (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 i}{2 \pi a} (\frac{2}{\sqrt{2}}) = \frac{\mu_0 i \sqrt{2}}{2 \pi a}$.For a square loop,the total magnetic field $B = 4 	imes B_1 = 4 	imes \frac{\mu_0 i \sqrt{2}}{2 \pi a} = \frac{2 \mu_0 i \sqrt{2}}{\pi a}$.Given $\mu_0 = 4 \pi 	imes 10^{-7} \ T \cdot m/A$,$i = 5 \ A$,and $a = 1 \ m$:$B = \frac{2 	imes (4 \pi 	imes 10^{-7}) 	imes 5 	imes \sqrt{2}}{\pi 	imes 1} = 40 \sqrt{2} 	imes 10^{-7} \ T$.Comparing this with $X \sqrt{2} 	imes 10^{-7} \ T$,we get $X = 40$.

The current of $5 \ A$ flows in a square loop of side $1 \ m$ placed in air. The magnetic field at the centre of the loop is $X \sqrt{2} \times 10^{-7} \ T$. The value of $X$ is . . . . . . .

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