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JEE Mains · Physics · STD 12 - 3. current electricity

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 \(\dots \; \pi A .\)

  1. A \(-48\)
  2. B \(48\)
  3. C \(-58\)
  4. D \(58\)
Verified Solution

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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