Question

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 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

The current is
`I = 8.9 *10^(-5) \ A`

Step-by-step explanation:

From the question we are told that

The radius is
`r = 3.17 \ mm = 3.17 *10^(-3) \ m`

The current density is
`J = c\cdot r^2 = 9.00*10^(6) \ A/m^4 \cdot r^2`

The distance we are considering is
`r = 0.5 R = 0.001585`

Generally current density is mathematically represented as

`J = (I)/(A )`

Where A is the cross-sectional area represented as

`A = \pi r^2`

=>
`J = (I)/(\pi r^2 )`

=>
`I = J * (\pi r^2 )`

Now the change in current per unit length is mathematically evaluated as

`dI = 2 J * \pi r dr`

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

`I = 2\pi \int\limits^(0.5 R)_(0) {( 9.0*10^6A/m^4) * r^2 * r} \, dr`

`I = 2\pi * 9.0*10^(6) \int\limits^(0.001585)_(0) {r^3} \, dr`

`I = 2\pi *(9.0*10^(6)) [(r^4)/(4) ] | \left 0.001585} \atop 0}} \right.`

`I = 2\pi *(9.0*10^(6)) [ (0.001585^4)/(4) ]`

substituting values

`I = 2 * 3.142 * 9.00 *10^6 * [ (0.001585^4)/(4) ]`

`I = 8.9 *10^(-5) \ A`