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

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?

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niczorrrr 5 years 2021-08-10T13:33:07+00:00 1 Answers 16 views 0

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  1. minhkhang
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    2021-08-10T13:34:47+00:00
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    Answer:

    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

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