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  4. A steady current is set up in a cubic network composed of wires of equal resistance and length d as shown in the figure. The magnetic field at the centre P due to the-cubic network is (n μ0/4 π) ⋅ (2 I/d). Find the value of n.

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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 the figure. The magnetic field at the centre $P$ due to the-cubic network is $\frac{n \mu_{0}}{4 \pi} \cdot \frac{2 I}{d}$. Find the value of $n$.Physics Question Image

Moving Charges and Magnetism

Solution:

The magnetic field at centre $P$ is zero.
The field due to $A B$ is cancelled by field due to $G F$.
Similarly, all the diagonally opposite wires cancel field of each other.
Therefore, $n=0$.