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 Ω m) (a) 11 × 10 - 5 V/m (b) 11 × 10 - 3 V/m (c) 11 × 10 - 7 V/m (d) 11 × 10 - 2 V/m

Answer: (a). I = J · A = JAcos 60° J = (5)/(0.04 × 0.5) = 250 A/m² E = ρ J = 44 × 10⁻⁸ × 250 = 1.1 × 10⁻⁴ = 11 × 10⁻⁵ V/m So the answer is (a).

A current of 5 A is passing through a non-linear magnesium wire of… | JEE Main 2021 (20 Jul, Shift 1)

Q24JEE Main 2021 · 20 JUL · SHIFT 1Medium

Electric Current, Drift Velocity & Current Density

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 $^\circ$ with the unit vector of area of cross- section. The magnitude of electric field at every point of the conductor is:

(Resistivity of magnesium $\rho$ = 44 $\times$ 10 $-$ 8 $\Omega$ m)

Show answer and solution

Answer: (a)

$I = \vec{J} \cdot \vec{A} = JA\cos 60^\circ$
$J = \dfrac{5}{0.04 \times 0.5} = 250\text{ A/m}^{2}$
$E = \rho J = 44 \times 10^{-8} \times 250 = 1.1 \times 10^{-4} = 11 \times 10^{-5}\text{ V/m}$
So the answer is (a).

Report a mistake

Electric Current, Drift Velocity & Current Density question from JEE Main 2021 (20 Jul, Shift 1)

More questions on Electric Current, Drift Velocity & Current Density