A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m². At every point the direction of current density is at an angle of 60° with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is :
(Resistivity of magnesium ρ = 44 × 10⁻⁸ Ωm)

Solution

Physics sir · 3 steps

Current is the flux of J⃗, so only the part of J along the area normal carries the 5 A.

1. Current as flux. Current through a surface is I = ∫ J⃗ · dA⃗. Here J⃗ makes 60° with the area normal at every point, so I = JA cos 60°.

   J = I/(A cos 60°) = 5/(0.04 × 0.5) = 250 A m⁻²

2. Ohm's law, local form. At any point of the conductor, E⃗ = ρ J⃗. So E = 44 × 10⁻⁸ × 250 = 1.1 × 10⁻⁴ V m⁻¹.
3. So. E = 11 × 10⁻⁵ V/m, which is option (a).

The idea to keep. When J⃗ is tilted from the area normal, the current is I = JA cos θ, not JA. Find the full J first, then use E = ρ J.

Answer from the NTA final answer key. Solution by PrepWiser.