The current density in a cylindrical wire of radius 4 mm is 4 $\times$ 106 Am$-$2. The current through the outer portion of the wire between radial distances ${R \over 2}$ and R is ____________ $\pi$ A.
Answer (integer)
48
Solution
<p>$i = A \times j$</p>
<p>$= \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$</p>
<p>$= {{3\pi {R^2}} \over 4} \times j$</p>
<p>$= {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 4 \times {10^6}$</p>
<p>$= 48\pi$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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