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

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 ..........

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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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