JEE PYQ: Current Electricity Question 15
Question 15 - 2021 (24 Feb 2021 Shift 2)
A cylindrical wire of radius 0.5 mm and conductivity $5 \times 10^7$ S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be $x^2\pi$ mA. The value of $x$ is
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Answer: 5
Solution
$J = \sigma E = 5 \times 10^7 \times 10 \times 10^{-3} = 5 \times 10^5$ A/m$^2$. $I = J \times \pi R^2 = 5 \times 10^5 \times \pi \times (0.5 \times 10^{-3})^2 = 5 \times 10^5 \times \pi \times 0.25 \times 10^{-6} = 125 \times 10^{-3}\pi = 25\pi$ mA $= 5^2\pi$ mA. $x = 5$.
BETA
