Q 12-03-176JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
The current density in a cylindrical wire of radius $4$ mm is $4\times10^6\ \text{A m}^{-2}$. The current through the outer portion of the wire between radial distances $\dfrac R2$ and $R$ is ______ $\pi$ A.
Numerical value type. Enter your answer.
Answer: 48
With uniform current density,
$$I = J\cdot\pi\left(R^2 - \frac{R^2}{4}\right) = J\cdot\frac{3}{4}\pi R^2$$
$$I = 4\times10^6\times\frac34\times\pi\times(4\times10^{-3})^2 = 4\times10^6\times12\times10^{-6}\,\pi = 48\pi\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics