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

`i = 4.34 * 10^(-5) A`

Step-by-step 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 = (\pi c)/(2) (r^4)`

now we have to find current in r = 0.5 R

so we have

`i = (\pi c)/(2)(0.5 R)^4`

now plug in all data

`i = (\pi(5.25 * 10^6))/(2)(0.5 * 3.03 * 10^(-3))^4`

`i = 4.34 * 10^(-5) A`