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A cylindrical conductor of diameter 0.1 mm carries a current of 90 mA. The current density( in Am^(-2)) is pi= 3 |
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Answer» `1.2xx10^(7)` Radius (R ) `= (D)/(2)=(0.1)/(2)mm` Current (I) = 90 mA `= 90xx10^(-3)A` We know that, Current DENSITY `(J)=(I)/(A)` `= (I)/(pi r^(2))` `= (90xx10^(-3))/((22)/(7)xx((0.1xx10^(-3))/(2))^(2))` `= (90xx10^(-3))/(3.14xx((0.1xx10^(-3))/(2))^(2))` `= 12000xx10^(3)` `= 1.2xx10^(7)A//m^(2)` |
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