A current is uniformly distributed across the cross-section of cylindrical wire of radius \( R = 4\,\mathrm{mm} \) with current density \( J = 2.5\times10^{5}\,\mathrm{A\,m^{-2}} \). The amount of current flowing through the outer portion of the wire between radial distance \( R/2 \) and \( R \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 9.42\,\mathrm{A} \)
For a uniform current density, the current through any part of the cross-section is \( I = J \times \) (area of that part). The part between \( R/2 \) and \( R \) is a ring-shaped region (annulus).
\( \displaystyle A = \pi R^{2} - \pi\left(\frac{R}{2}\right)^{2} = \frac{3}{4}\,\pi R^{2} \)
\( \displaystyle R = 4\,\mathrm{mm} = 4\times10^{-3}\,\mathrm{m} \;\Rightarrow\; R^{2} = 1.6\times10^{-5}\,\mathrm{m^{2}} \)
\( \displaystyle A = \frac{3}{4} \times \pi \times 1.6\times10^{-5} = 12\pi\times10^{-6}\,\mathrm{m^{2}} \)
\( \displaystyle I = J\,A = 2.5\times10^{5} \times 12\pi\times10^{-6} = 3\pi = 9.42\,\mathrm{A} \)