Sankalp NEET Full Test-4 Question-26 Solution

Question: 26: A copper wire of length 10 m and radius \left(\frac{10^{-2}}{\sqrt{\pi}}\right) \mathrm{m} has electrical resistance of 10 \Omega. The current density in the wire for an electric field strength of 10(\mathrm{~V} / \mathrm{m}) is

(1) 10^{-5} \mathrm{~A} / \mathrm{m}^{2}

(2) 10^{5} \mathrm{~A} / \mathrm{m}^{2}

(3) 10^{4} \mathrm{~A} / \mathrm{m}^{2}

(4) 10^{6} \mathrm{~A} / \mathrm{m}^{2}

Answer: Option (2)

Explanation:

The area of cross-section of the wire is given by

A = \pi r^2

Substituting the value of radius,

A = \pi \left(\frac{10^{-2}}{\sqrt{\pi}}\right)^2 A = \pi \cdot \frac{10^{-4}}{\pi} = 10^{-4} \ \mathrm{m}^2

The resistance of a wire is related to its resistivity by

R = \frac{\rho L}{A}

Rearranging, the resistivity is

\rho = \frac{R A}{L}

Substituting the given values,

\rho = \frac{10 \times 10^{-4}}{10} = 10^{-4} \ \Omega \mathrm{m}

The current density J is related to the electric field E by

J = \frac{E}{\rho}

Substituting the values,

J = \frac{10}{10^{-4}} = 10^{5} \ \mathrm{A} / \mathrm{m}^2

Hence, the correct answer is Option (2).