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Question

(A) Consider a square frame $ABCD$ with side length $a$. When a battery is connected across a diagonal (say $A$ and $C$),the current $I$ entering at $

Vedclass · Accepted Answer

Zero

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(A) Consider a square frame $ABCD$ with side length $a$. When a battery is connected across a diagonal (say $A$ and $C$),the current $I$ entering at $A$ splits into two equal parts,$I/2$,flowing through paths $ABC$ and $ADC$.For the path $ABC$,the current $I/2$ flows through segments $AB$ and $BC$. By the Biot-Savart Law,the magnetic field produced by segment $AB$ at the center $O$ is directed into the plane of the paper. Similarly,the field due to segment $BC$ is also directed into the plane.For the path $ADC$,the current $I/2$ flows through segments $AD$ and $DC$. The magnetic field produced by segment $AD$ at the center $O$ is directed out of the plane of the paper. Similarly,the field due to segment $DC$ is also directed out of the plane.Since the segments are symmetric and the currents are equal,the magnetic field due to $AB$ and $BC$ (inward) is exactly cancelled by the magnetic field due to $AD$ and $DC$ (outward).Therefore,the net magnetic field at the center of the square is $0$.

On connecting a battery to the two corners of a diagonal of a square conductor frame of side $a$,the magnitude of the magnetic field at the centre will be:

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