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Question

(B) The total current $I$ flowing through the wire is given by the integral of current density $J$ over the cross-sectional area $A$: $I = \int J \, d

Vedclass · Accepted Answer

$37 \pi 	ext{ kJ}$

Vedclass · Answer

(B) The total current $I$ flowing through the wire is given by the integral of current density $J$ over the cross-sectional area $A$: $I = \int J \, dA$. Since the wire is circular,$dA = 2 \pi r \, dr$. $I = \int_0^{R} J(r) \cdot 2 \pi r \, dr$,where $R = 2 	imes 10^{-3} 	ext{ m}$. $I = \int_0^{2 	imes 10^{-3}} (10^5 r) \cdot (2 \pi r) \, dr = 2 \pi 	imes 10^5 \int_0^{2 	imes 10^{-3}} r^2 \, dr$. $I = 2 \pi 	imes 10^5 \left[ \frac{r^3}{3} ight]_0^{2 	imes 10^{-3}} = 2 \pi 	imes 10^5 	imes \frac{8 	imes 10^{-9}}{3} = \frac{16 \pi}{3} 	imes 10^{-4} 	ext{ A}$. The energy consumed $E$ is given by $E = V \cdot I \cdot t$. $E = 70 	ext{ V} 	imes \left( \frac{16 \pi}{3} 	imes 10^{-4} 	ext{ A} ight) 	imes 1000 	ext{ s}$. $E = 70 	imes \frac{16 \pi}{3} 	imes 10^{-1} = \frac{1120 \pi}{30} = \frac{112}{3} \pi \approx 37.33 \pi 	ext{ J}$. Given the options,the closest value is $37 \pi 	ext{ J}$ (Note: The unit in the provided options appears to be in Joules,not kJ).

The current density in a circular wire is given by $J(r) = (1 \times 10^5 \text{ A/m}^3) r$,where $r$ is the radial distance and the wire's radius is $2 \text{ mm}$. If the potential applied across the wire is $70 \text{ V}$,then the energy consumed by the wire in $1000 \text{ s}$ is:

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