The magnitude J of the current density in a certain wire with a circular cross section of radius R = 3.50 mm is given by J = (5.00 ✕ 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 = 0.880R and r = R?

0.0472A or 47.2mA

Explanation:

Given

J = 5.00 * 10^8 r²

Radius, R = 3.50mm

Boundary; r = 0.880R and r = R

The question asks for the current through the area bounded by R and 0.880R.

The area can be calculated by the integral below;

∫J.dA {0.880R,R} where dA = r²2πrdr

= ∫j * r²2πr dr

= ∫2πJr³ dr {0.880R,R}

= 2πJ∫r³ dr {0.880R,R}

Integrate...

= 2πJ[r⁴/4] {0.880R,R}

= ½πJ[r⁴] {0.880R,R}

= ½πJ(R⁴ - (0.880R)⁴)

Substitute J = 5.00 * 10^8 r² and, R = 3.50mm

R = 3.5 * 10^-3 m

= ½ * 5.00 * 10^8 * π ((3.5 * 10^-3)⁴ - (0.880 * 3.5 * 10^-3)⁴)

= ½ * π * 5 * 10^8 * 6.01 * 10^-11

= 0.047202429620186

= 0.0472A or 47.2mA

Hence, the current through the outer section bounded by r = 0.880R and r = R is 0.0472A or 47.2mA

33: i think it’s b

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.