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

Correct Answer: 48

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

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

\(= \frac {3π \times (4 \times 10^{-3})^2}{4} \times 4 \times 10^6\)
\(= 48 \pi\)
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 \(48\).