Recent content by FatoonsBaby71

Telling from the answer it looks like aphi

Help with curls and current density!1 Homework Statement In cylindrical coordinates, J = 105*(cos(2r)2)*az in a certain region. Obtain H from this current density and then take the curl of H and compare with J. Answer H = 105*(r/4+sin4r/8+cos4r/32r-1/32r)aphi Homework Equations J =...

No unfortunately the answer is in milli. I don't know why... I may have to stick to it though...

Oh, I see ... Now I understand. We take the total area and take away the hollow part leaving us with the filled area... then plugging that into the equation gets us the result... Thanks Doc you were such a good help.. I hope you don't mind...I don't want to bug you...but could you provide...

Okay so i understand what your are saying...So the set up should look like this... 1/(38.2*10^6*(pi*(6^2))) = 2.3146*10^-10 ? Which does not even look remotely close to the answer. From what you stated it looks like the empty part is from 0 to 10 and then the radius in total is 16 so...

I was actually doing this from the start...because i changed the three and five to meters, I still get the -9 power answer? its not the ln part of the equation its the other part??

I see the reason why the length is not needed and you can use 1. But in this case is the area... just half of 32 for the radius which is 16mm and since the wall thickness is 6m i would want to take the area from 0 to 10 mm?

Help with conductors! Homework Statement Determine the resistance per meter of a hollow cylindrical aluminum conductor with an outer diameter of 32mm and wall thickness 6mm. Homework Equations R = length / sigma * Area Sigma = conducitivity The Attempt at a Solution The answer is...

Help with resistance! Homework Statement Determine the resistance of a copper conductor 2m long with a circular cross section and a radius of 1mm at one end increasing linearly to a radius of 5mm at the other. Homework Equations R = ( 1 / 2pi * sigma * length ) * ln (b/a) sigma =...

Yes, I was wrong the uniform line charge is in the x-y plane where z = 0... I apologize. So I really can't just pick a point on the line charge...i thought it wouldn't matter because it is uniformly distributed. Do you have any idea on how I could attack this problem??

Homework Statement A uniform line charge of density pl = 1nC/m is arrange in the form of a square 6m on a side. (In the z plane) Find the potential at (0,0,5) m Answer:35.6 V Homework Equations R = The distance from the line to the point (0,0,5) dL = ax dx + ay dy + az dz The...

\vec{D} = 2xy\hat{i} + x^2\hat{j} is correct. When you say that several are zero flux, I assume your referring to the flux in the z direction because there is no z component. Another thing that points out is the fact that the y component is x^2. Due to symmetry, I am assuming that both fluxes...

Oh ok that makes sense. However, I wanted to check if I computed the charge correctly. I had two double integrals... The first double integral went from 0 to 3 then 0 to 2 of (2xy) * dy dz which resulted in 17 C The second double integral went from 0 to 3 then 0 to 1 of x^2 * dx dz which...

Homework Statement The electric flux density in space is given as D = 2xy(ax)+x^2(ay) C/m^2. A box is given as: 0<=x<=1; 0<=y<=2;0<=z<=3 (m). It is found that the total electric flux out of the box is (Wo). The charges creating the electric flux density D are now removed and replaced by a...