The magnitude J of the current density in a wine with a circ

The magnitude J of the current density in a wine with a circular cross section of radius R = 3.00 mm is given by J = (4.00 x 10^8)r^3. with J in amperes per square meter and radial distance r in meters Determine the current through the outer section bounded by r - 0.800 R and r=R.

Solution

Since J is dependent on radial distance, there\'s going to be integration involved.

In the circumstance of a uniform current density, we would have I = JA. Since the current density varies, we must write this instead dI = J dA. Since the density is radially symmetric, it is easiest to work in cylindrical coordinates, in which case we get dA = r dr d, and since J does not depend on angular displacement, i.e. it only depends on distance and not angle, this becomes r dr d, = 0 to 2 = 2r dr. Hence, dI = 2rJ dr.

I = 2rJ dr, r = 0.800R to R
= 2r(4.00 x 10^8 r²) dr, r = 0.800R to R
= (8 x 10^8) r³ dr, r = 0.800R to R
= (8 x 10^8) (r/4, r = 0.800R to R)
= (2 x 10^8) (R - (0.800R))
= (2R x 10^8)(1 - (0.800)) = 2.48 mA

You can then sub in the value R = 2.40 mm to obtain I = 8.57 mA [1]. Notice that for J to have units of A/m², the constant 3 x 10^(8) must have units A/m.