The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 $\times$ 106 A/m2. The current through the outer portion of the wire between radial distances ${r \over 2}$ and r is x$\pi$ A; where x is __________.
Answer (integer)
12
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>$$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 1.0 \times {10^6}$$</p>
<p>$= 12\,\pi$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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