Question

The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.11 mm is given by J = (3.25 × 108)r2, with J in amperes per square meter and radial distance r in meters. What is the current through the outer section bounded by r = 0.921R and r = R?

Answer 1

Answer:

$i = 2.84 \times 10^{-3} A$

Explanation:

As we know that current density is ratio of current and area of the crossection

now we have

$J = \frac{di}{dA}$

so the current through the wire is given as

$i = \int J dA$

now we have

$i = \int_{0.921R}^R J dA$

here we have

$J = (3.25 \times 10^8)r^2$

now plug in the values in above equation

$i = \int_{0.921R}^R (3.25 \times 10^8)r^2 2\pi r dr$

now we have

$i = \int_{0.921R}^R 2\pi (3.25 \times 10^8)r^3 dr$

$i = (2.04 \times 10^9) \frac{r^4}{4}$

now plug in both limits as mentioned

$i = (2.04 \times 10^9)(\frac{R^4}{4} - \frac{(0.921R)^4}{4})$

$i = (2.04\times 10^9)(0.07 R^4)$

here R = 2.11 mm

$i = (2.04 \times 10^9)(0.07 (2.11 \times 10^{-3})^4)$

$i = 2.84 \times 10^{-3} A$