A current of 4 A is passed through a square | vedclass

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(D) The current enters at one corner and leaves at the adjacent corner of the square loop.This divides the loop into two parallel paths: one path cons

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(D) The current enters at one corner and leaves at the adjacent corner of the square loop.This divides the loop into two parallel paths: one path consists of one side of the square (length $l = 5 \ cm$),and the other path consists of three sides of the square (length $3l = 15 \ cm$).Since the wire is uniform,the resistance of the paths is proportional to their lengths. Let $R$ be the resistance of one side. Then the resistance of the first path is $R_1 = R$ and the second path is $R_2 = 3R$.The current $I = 4 \ A$ splits into $I_1$ and $I_2$ such that $I_1 R_1 = I_2 R_2$,which means $I_1 R = I_2 (3R)$,so $I_1 = 3I_2$.Since $I_1 + I_2 = 4 \ A$,we get $4I_2 = 4 \ A$,so $I_2 = 1 \ A$ and $I_1 = 3 \ A$.The magnetic field at the center due to a straight wire segment of length $L$ carrying current $I$ at a perpendicular distance $a$ is $B = \frac{\mu_0 I}{4 \pi a} (\sin 	heta_1 + \sin 	heta_2)$.For a square of side $a = 5 \ cm = 0.05 \ m$,the distance from the center to each side is $d = a/2 = 2.5 \ cm = 0.025 \ m$.For each side,$	heta_1 = 	heta_2 = 45^\circ$,so $\sin 45^\circ + \sin 45^\circ = \sqrt{2}$.The magnetic field due to one side carrying current $I$ is $B = \frac{\mu_0 I}{4 \pi (a/2)} (\sqrt{2}) = \frac{\mu_0 I \sqrt{2}}{2 \pi a}$.The magnetic field produced by the path with $I_1 = 3 \ A$ (one side) is $B_1 = \frac{\mu_0 (3) \sqrt{2}}{2 \pi (0.05)}$.The magnetic field produced by the path with $I_2 = 1 \ A$ (three sides) is $B_2 = 3 	imes \frac{\mu_0 (1) \sqrt{2}}{2 \pi (0.05)}$.Since the currents flow in opposite directions around the center,the fields $B_1$ and $B_2$ are in opposite directions.Thus,the net magnetic field is $B_{net} = |B_1 - B_2| = |\frac{3 \mu_0 \sqrt{2}}{2 \pi (0.05)} - \frac{3 \mu_0 \sqrt{2}}{2 \pi (0.05)}| = 0$.

$A$ current of $4 \ A$ is passed through a square loop of side $5 \ cm$ made of a uniform manganin wire as shown in the figure. The magnetic field at the centre of the loop is

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