Question

The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.70 mm is given by J = (5.50 ✕ 108)r2, with J in amperes per square meter and radial distance r in meters. What is the current through the outer section bounded by r = 0.510R and r = R?

Answer 1

given R = 2.70 mm

current density J = (5.50 ✕ 10⁸)r²

current through the outer section bounded by r = 0.510R and r = R

we know that

i = 2 π r J dr
= 2π (5.50 ✕ 10⁸) r³ dr
= 11π x 10⁸ r³ dr
= 11π x 10⁸   ( r⁴ / 4 ) r = 0.510R t R
= 2.75π x 10⁸   ( R⁴ - (0.510R)⁴ )
= (2.75 π R⁴ X 10⁸ )( 1 - (0.510)⁴ )

R = 2.70 mm = 0.27 cm = 0.0027m

i = 2.75 X 3.14 X 0.0027⁴ ( 1 - (0.510)⁴ )

i = 4.58 X 10^-10 ( 1 - 0.0676)

i = 4.58 X 10^-10 X 0.9324

i = 4.27 X 10^-10 A