The magnitude J of the current density in a certain wire wit

The magnitude J of the current density in a certain wire with a circular cross section of radius R = 2.15 mm is given by J = (3.34 × 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 = 0.912R 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 = 0.912R to R
= 2r(3.34 x 10^8 r²) dr, r = 0.912R to R
= (6.68 x 10^8) r³ dr, r = 0.912R to R
= (6.68 x 10^8) (r/4, r = 0.912R to R)
= (1.67 x 10^8) (R - (0.912R))
= (1.67R x 10^8)(1 - (0.912))
Putting the value of R=2.15 mm and =3.14,we get the value of current I,

=3.45x 10^-3 Amperes or 3.45 milliAmperes(mA).(Ans)