JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 24)
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
12
Explanation
$$i = A \times j$$
$$ = \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$$
$$ = {{3\pi {R^2}} \over 4} \times j$$
$$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 1.0 \times {10^6}$$
$$ = 12\,\pi $$
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