Question

A current of 5A is passing through a non-linear magnesium wire of cross-section 0.04m². 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 = 44 × 10^-8 Ωm

पर्याय

• 11 × 10^-3 V/m

• 11 × 10^-5 V/m

• 11 × 10^-7 V/m

• 11 × 10^-2 V/m

Answer 1

A current of 5A is passing through a non-linear magnesium wire of cross-section 0.04m². 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 11 × 10^-5 V/m.

Explanation:

Electric current (I) = 5A

Cross-section area (A) = 0.04 m²

Angle (θ) = 60°

Resistivity (ρ) = 44 × 10^-8 ωm

Current density (J) = `1/(Acostheta)`

= `(5A)/(0.04 cos60^circ) = 250` A/m²

The magnitude of electric field 'E' = ρJ

= 44 × 10^-8 × 250

= 11 × 10^-5 V/m