GATE 2025 EC Q.50: solution and answer

Official answer: ASource: Official answer key, GATE 2025 EC (IIT Roorkee)

1. Ohm's law in point form: the conduction current density is $\mathbf{J}=\sigma \mathbf{E}$. It points along the wire, in the same direction as $\mathbf{E}$.

2. $$J=\sigma E=(5.8\times 10^7\text{S/m})(0.01\text{V/m})=5.8\times 10^5{\text{A/m}}^2$$

3. Cross-section: the diameter is $1\text{mm}$, so the radius is $r=0.5\text{mm}=5\times 10^{-4}\text{m}$ and $$A=\pi r^2=\pi (5\times 10^{-4}{)}^2=7.854\times 10^{-7}{\text{m}}^2$$

4. The field is steady (DC), so there is no skin effect and $\mathbf{J}$ is uniform over the cross-section. The current is therefore $I={\int }_S\mathbf{J}\cdot d\mathbf{S}=JA$.

5. $$I=(5.8\times 10^5)(7.854\times 10^{-7})=0.4555\text{A}\approx 0.46\text{A}$$

6. This is option (A).