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 \(\dots \; \pi A .\)
- A \(-48\)
- B \(48\)
- C \(-58\)
- D \(58\)
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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