Question

A straight, cylindrical wire lying along the x axis has a length of 0.616 m and a diameter of 0.6 mm. It is made of a material described by Ohm's law with a resistivity of ρ = 4.72 ✕ 10−8Ω· m. Assume a potential of 4.85 V is maintained at the left end of the wire at x = 0. Also assume V = 0 at x = 0.616 m. Find the current density (in A/m2) in the wire.

Answer 1

Resistivity of the wire = = 4.72 x 10^-8 . m.

Length of the wire = L = 0.616 m.

Area of its cross - section be A, in m².

Hence, resistance of the wire = R = L / A.

Potential difference applied to its ends = V = 4.85 V ( since V = 0 V at the other end ).

Let, the current in the wire be I, in A.

Hence, according to Ohm's law,

V = IR = I x L / A

or, V = ( I / A ) x L

or, I / A = V / L.

But, we know, I / A is called current density, i.e., current per unit area. Let us denote this by J.

Hence, J = V / L = [ 4.85 / ( 4.72 x 10^-8 x 0.616 ) ] A / m²

or, J ~ 1.67 x 10⁸ A / m².