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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2022
  • Medium
  • Numerical

The current density in a cylindrical wire of radius \(4 \; mm\) is \(4 \times 10^{6} \; Am ^{-2}\). The current through the outer portion of the wire between radial distance \(\frac{R}{2}\) and \(R\) is \(\dots \; \pi A .\)

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Answer & Solution

Correct Answer

(B) \(48\)

Step-by-step Solution

Detailed explanation

\(J=\frac{I}{A}\)

\(I=J A\)

\(=4 \times 10^{6} \times\left[\pi R^{2}-\pi\left(\frac{R}{2}\right)^{2}\right]\)

\(=4 \times 10^{6} \times \pi R^{2} \times \frac{3}{4}\)

\(=4 \times 10^{6} \times \pi \times\left(4 \times 10^{-3}\right)^{2} \times \frac{3}{4}=48 \; \pi A .\)

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