A steady current is set up in a cubic network | vedclass

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(C) The cubic network consists of $12$ identical wires,each of resistance $R$ and length $d$. Due to the symmetry of the cube and the arrangement of t

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$0$

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(C) The cubic network consists of $12$ identical wires,each of resistance $R$ and length $d$. Due to the symmetry of the cube and the arrangement of the current flow,for every current-carrying wire segment,there exists a diametrically opposite wire segment carrying an equal current in the opposite direction relative to the center $P$. According to the Biot-Savart law,the magnetic field produced by these symmetric segments at the center $P$ cancels each other out. Therefore,the net magnetic field at the center $P$ of the cubic network is $0$.

$A$ steady current is set up in a cubic network composed of wires of equal resistance and length $d$ as shown in the figure. What is the magnetic field at the centre $P$ due to the cubic network?

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