Question

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 center $P$ due to the cubic network?

• A. $\frac{{\mu }_0}{4\pi }\frac{2I}{d}$
• B. $\frac{{\mu }_0}{4\pi }\frac{3}{\sqrt{2}d}$
• C. $0$
• D. $\frac{{\mu }_0}{4\pi }\frac{\theta \pi I}{d}$

Answer 1

Answer: (C)

$0$

All the diagonally opposite arms will carry equal current.
Net magnetic field at $P$ will be zero, because the magnetic field at center $P$ is canceled by opposite pairs.
Magnetic field at P due to AD and EF will cancel each other.
Magnetic field at P due to HG and BC will cancel each other.
Magnetic field at P due to AB and GF will cancel each other.
Magnetic field at P due to DC and HE will cancel each other.
Magnetic field at P due to EB and GD will cancel each other.
Magnetic field at P due to HA and FC will cancel each other.