Current I is flowing along the path ABCDA con | vedclass

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(D) In figure $-a$,the path $ABCDA$ forms a square loop in the $XZ$-plane. The magnetic field $B_0$ at the center due to this loop is directed along t

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$\sqrt{3} B_0$ directed towards corner $F$

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(D) In figure $-a$,the path $ABCDA$ forms a square loop in the $XZ$-plane. The magnetic field $B_0$ at the center due to this loop is directed along the $Y$-axis.In figure $-b$,the path $ABCGHEA$ can be viewed as the superposition of three square loops in the $XY$,$YZ$,and $XZ$ planes,each carrying current $I$.The magnetic field produced by each square loop of current $I$ at the center of the cube has a magnitude $B_0$ and is directed along the axis perpendicular to the plane of the loop.Thus,the total magnetic field $B$ at the center is the vector sum of the fields produced by these three loops: $\vec{B} = \vec{B}_x + \vec{B}_y + \vec{B}_z$.Since the magnitudes are equal,$B = \sqrt{B_0^2 + B_0^2 + B_0^2} = \sqrt{3} B_0$.The direction of the resultant field is along the diagonal of the cube,which points towards corner $F$.

Current $I$ is flowing along the path $ABCDA$ consisting of four edges of a cube (figure $-a$),producing a magnetic field $B_0$ at the centre of the cube. Find the magnetic field $B$ produced at the center of the cube by a current $I$ flowing along the path of the six edges $ABCGHEA$ (figure $-b$).

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