Hello guys. I gave this one the good ol effort and seem to have gotten stuck. Can anyone help? ----------- A copper wire with a circular cross-section area of 1.5 mm2 carries a current of 11 A. The resistivity of the material is 2.1 × 10-8 ·m. a) What is the uniform electric field in the material? b) If the current is changing at the rate of 4800 A/s, at what rate is the electric field in the material changing? c) What is the displacement current density in the material in part (b)? (Hint: Since K for copper is very close to 1, use =0.) d) If the current is changing as in part (b), what is the magnitude of the magnetic field 8.8 cm from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of B. Is the contribution from the displacement current significant? Okay I solved a) b) and c) a) .154v/m b)67.2v/M*s c)5.94989E-10 A/m^2 now trying to get the last part I got id=(jd)A id=(5.94989E-10)*(1.5E-6)=8.92483E-16 And generalized version of Amepre's law states Line integral of B*dl=mu-0(ic+id) ic=(dE/dt)*(A)*(e-0)=8.92483E-16 So the Bc part is Bc=(ic/(2*pi*.088m)) = 2.0283E-21 And I got the same thing for Bd Confused Anyone???????? THANKS!!!!!!!!