Question

9. The current i in a wire of circular cross section can be calculated from the current density, J, using the following equation: i =  J dA. Given that the magnitude of the current density, J = (3 x 108 ) r2 amps/m2 For a wire with a radius of R = 2.00 mm what is the current in the outer section from r = 0.90 R to r = R?

Answer 1

Answer:

Explanation:

i = ∫J dA

A = πr²

dA = 2πr dr

i = ∫J 2πr dr

=  ∫3 x 10⁸ x  2πr³ dr

=3 x 10⁸ x  2π  ∫ r³ dr

integrating and taking limit from r = .9 R to R

3 x 10⁸ x  2π / 4 [ 2⁴ - .9⁴ x  2⁴] x (10⁻³)⁴

= 4.71  x 10⁸ x .3439 x 10⁻¹² x 2⁴

= 25.9  x 10⁻⁴ A