A copper wire of length 10 m and radius \(\left(10^{-2} / \sqrt{\pi}\right) \) m has electrical resistance of 10Ω . The current density in the wire for an electric field strength of 10 (V/m) is:

CONCEPT:

The current density is defined as the product of conductivity and electric field and it is written as;

J = σE

Here, σ  is the conductivity and E is the electric field.

CALCULATION:

Given:

Length, l = 10 m

radius, r = \(\left(10^{-2} / \sqrt{\pi}\right) \)

and Electric resistance, R = 10 Ω 

E = 10 V/m

According to the current density we have;

J = σE     -----(1)

The resistivity is defined as the reciprocal of the conductivity, therefore, in equation (1) we have;

J = \(\frac{1}{ρ }\) E    -----(2)

Now, resistivity, ρ  =\(\frac{RA}{l}\)

Here, R is the resistance, A is the area of cross-section, and l is the length. Putting these values of resistivity in equation (2) we have;

J = \(\frac{l}{RA }\) E    ----(3)

The area of cross-section ,A = \(\pi r^2\) = 10^-4 m²

Now, on putting all the given values in equation (3) we have;

J = \(\frac{10}{10 \times 10^{-4} } \times 10\) 

⇒J = 10⁵ A/m²

Hence, option 1) is the correct answer.