JEE MAIN - Physics (2022 - 27th June Morning Shift - No. 23)
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
48
Explanation
$$i = A \times j$$
$$ = \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j$$
$$ = {{3\pi {R^2}} \over 4} \times j$$
$$ = {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 4 \times {10^6}$$
$$ = 48\pi $$
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