Q 12-03-201JEE MainJEE Main 2021 (20 Jul, Shift 1)Medium
A current of 5 A is passing through a non-linear magnesium wire of cross-section $0.04\ \text{m}^2$. At every point the direction of current density is at an angle of $60^\circ$ with the unit vector of area of cross-section.
The magnitude of electric field at every point of the conductor is: (resistivity of magnesium $\rho = 44\times10^{-8}\ \Omega$ m)
Answer: (C) $11\times10^{-5}\ \text{V m}^{-1}$
$I = JA\cos60^\circ \Rightarrow J = \dfrac{5}{0.04\times0.5} = 250\ \text{A m}^{-2}$.
$$E = \rho J = 44\times10^{-8}\times250 = 1.1\times10^{-4} = 11\times10^{-5}\ \text{V m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics