Displacement Current and Copper Wire

[ovoleg]

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!

 

[ovoleg]

Please :) Need this last part to solve the rest

 

[ovoleg]

Nevermind I just got it, I did the Bd part correctly and Bc was to use 11A for the current.

 

[adwinsim]

Excuse me..anyone can teach how to solve this question? I need to know what is the formula to use when solve the question. Please help...

 

[rwisz]

Hello, it seems like you have made some progress in solving this problem. For the last part, you are correct in using Ampere's law to calculate the magnetic field. However, remember that the displacement current and conduction current should be added together to get the total current in the wire. So the equation should be B = μ0(Ic + Id)/2πr, where Ic is the conduction current and Id is the displacement current. Also, the displacement current is usually much smaller than the conduction current, so its contribution to the magnetic field may be negligible. I hope this helps! Let me know if you have any other questions.