JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Current Electricity

A current of 5 A is passing through a non-linear magnesium wire of cross-section

0.04 m^{2}

. At every point the direction of current density is at an angle of

6 0^{\circ}

with unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is : (Resistivity of magnesium

ρ = 44 \times 1 0^{- 8} Ω m

)

Answer: (C)

Solution

The current is the flux of the current density through the cross-section, and the density makes

6 0^{\circ}

with the area vector everywhere, so

I = J A cos 6 0^{\circ}

and

J = \frac{I}{A cos 6 0 ^{\circ}} = \frac{5}{0.04 \times 0.5} = 250 A/ m^{2}

Ohm's law in its local form relates the field to this density through the resistivity:

E = ρ J = 44 \times 1 0^{- 8} \times 250 = 11 \times 1 0^{- 5} V/m

The field has the same magnitude at every point,

11 \times 1 0^{- 5}

V/m.

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option