Physics, 12.08.2020 06:01 aaron0828

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?

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Answer from: alyssadgonzalez

The current is $I = 8.9 *10^{-5} \ A$

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

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

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

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

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

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

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

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

Answer from: Quest

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Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the curre...

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