A metallic rod has a length 1.5m and a diameter 0.2 cm. The rod carries a current of 5A when a p. d of 75V is applied between its ends.

(a) Find the current density in the rod.

(b) Calculate the magnitude of the electric field applied to the rod.

(c) Calculate the resistivity and conductivity of the material of the rod.

Explanation:

Current density J = i / A where i is current and A is cross sectional area

J = 5 / π x ( .1 x 10⁻² )²

= 1.6 x 10⁶ A / m²

b )

electric field applied E = V / l where V is potential difference and l is lemgth of rod

E = 75 / 1.5

= 50 v / m

c )

J = σ E

J is current density , σ is conductivity and E is electric field .

1.6 x 10⁶ = σ  x 50

σ = 3.2 x 10⁴ Ω⁻¹ m⁻¹

Resistivity = 1 / σ

= 1 / 3.2 x 10⁴

= .3125 x 10⁻⁴

=3.125 x 10⁻⁵  Ω m

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.