Question

Consider a wire of a circular cross-section with a radius of R = 3.03 mm. The magnitude of the current density is modeled as J = cr2 = 5.25 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?

Answer 1

Answer:

$i = 4.34 \times 10^{-5} A$

Explanation:

As we know that current in a wire depends on its current density and given as

$i = \int J.dA$

so we have

$i = \int cr^2 2\pi r dr$

so we will have

$i = 2\pi c \int r^3 dr$

$i = \frac{\pi c}{2} (r^4)$

now we have to find current in r = 0.5 R

so we have

$i = \frac{\pi c}{2}(0.5 R)^4$

now plug in all data

$i = \frac{\pi(5.25 \times 10^6)}{2}(0.5 \times 3.03 \times 10^{-3})^4$

$i = 4.34 \times 10^{-5} A$