Question

The current density in a cylindrical wire of radius 4 mm is $4×10^6\ Am^{–2}$. The current through the outer portion of the wire between radial distances $\frac R2$ and $R$ is _____ $\pi$ A.

Answer 1

$i = A \times j$
$= π (R_2 -\frac { R_2}{4})j$

$= \frac {3\pi R^2}{4} \times j$

$= \frac {3π \times (4 \times 10^{-3})^2}{4} \times 4 \times 10^6$
$= 48 \pi$
Let given that the current through the outer portion of the wire between radial distances $\frac R2$ and $R$ is $x\pi$ A.
Then, the value of $x = 48$.

So, the answer is $48$.