PQRS is a square loop made of uniform conduct | vedclass

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Question

(B) The square loop is divided into two parallel paths between points $P$ and $S$. One path is the side $PS$ itself, and the other path is the series

Vedclass · Accepted Answer

Zero at the centre of loop

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(B) The square loop is divided into two parallel paths between points $P$ and $S$. One path is the side $PS$ itself, and the other path is the series combination of sides $PQ$, $QR$, and $RS$.Since the wire is uniform, the resistance of each side is proportional to its length. Let the side length be $a$. The resistance of path $PS$ is $R_1 = \lambda a$, and the resistance of path $PQRS$ is $R_2 = 3\lambda a$, where $\lambda$ is the resistance per unit length.According to the Biot-Savart law, the magnetic field $B$ at the center due to a straight wire segment of length $L$ carrying current $I$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin 	heta_1 + \sin 	heta_2)$, where $d$ is the perpendicular distance from the center to the wire.For the square loop, the current splits at $P$ and recombines at $S$. The magnetic field produced by the path $PQRS$ at the center is the vector sum of the fields produced by segments $PQ$, $QR$, and $RS$. Due to symmetry and the direction of current flow, the magnetic field produced by the path $PQRS$ at the center is exactly equal and opposite to the magnetic field produced by the path $PS$ at the center.Therefore, the net magnetic field at the center of the loop is zero.

$PQRS$ is a square loop made of uniform conducting wire. If the current enters the loop at $P$ and leaves at $S$, then the magnetic field will be

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