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  • 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 the centre of cube P due to the cubic network ?</p

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 the centre of cube P due to the cubic network ?

Option: 1

\frac{\mu _{0}}{4 \pi}\frac{2I}{d}


Option: 2

\frac{\mu _{0}}{4 \pi}\frac{3I}{\sqrt{2}d}


Option: 3

0


Option: 4

\frac{\mu _{0}}{4 \pi}\frac{\theta \pi I}{d}


Answers (1)

best_answer

 

All the diagonally opposite arms will carry equal current.
Net magnetic field at P will be zero, because the magnetic fild 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.

 

By symmetry, the magnetic field at the centre P is zero

Posted by

Rishabh

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