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Q.

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

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a

$\frac{μ_{0}}{4 π} \frac{2 I}{d}$

b

$\frac{μ_{0}}{4 π} \frac{3 I}{\sqrt{2} d}$

c

0

d

$\frac{μ_{0}}{4 π} \frac{θ π I}{d}$

answer is C.

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Detailed Solution

Net magnetic field at l will be zero, because the magnetic field at centre P is cancelled by opposite pairs like AD-EF, etc.

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