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 ..........
- A \(10\)
- B \(11\)
- C \(12\)
- D \(14\)
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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