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.

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. 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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

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.

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 $\frac{n μ _{0}}{4 π} \cdot \frac{2 I}{d}$ . Find the value of $n$ .

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

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 4πnμ0​​⋅d2I​. Find the value of n.

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

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 $\frac{n μ _{0}}{4 π} \cdot \frac{2 I}{d}$ . Find the value of $n$ .

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