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

The current density in a cylindrical wire of radius \(r=4.0 \,mm\) is \(1.0 \times 10^{6} \,A / m ^{2}\). The current through the outer portion of the wire between radial distances \(r / 2\) and \(r\) is \(x \pi A\); where \(x\) is ..........

  1. A \(10\)
  2. B \(11\)
  3. C \(12\)
  4. D \(14\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(12\)

Step-by-step Solution

Detailed explanation

\(I =\int JdA\) \(=\int 10^{6} \times 2 \pi xdx\) \(\left.=10^{6} \times 2 \pi \cdot x \frac{ x ^{2}}{2}\right]_{\frac{ r }{2}}^{ T }\) \(=\pi \times 10^{6}\left[ r ^{2}-\frac{ r ^{2}}{4}\right]=12 \pi\) \(x =12\)
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