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 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 )

Question

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 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 ), 11 × 10 - 2 V m - 1, 11 × 10 - 7 V m - 1, 11 × 10 - 5 V m - 1, 11 × 10 - 3 V m - 1

MARKS App · Accepted Answer

I = J → · A → = J A cos ( θ ) 5 = J 4 100 × cos ( 60 ) J = 5 × 50 = 250 A m - 2 Now, E → = ρ × J → = 44 × 10 - 8 × 250 = 11 × 10 - 5 V m - 1

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