posted by Spencer on .
A 68.0 g hollow copper cylinder is 90.0 cm long and has an inner diameter of 1.0 cm. The current density along the length of the cylinder is 1.60×10^5(A/m^2). What is the current in the cylinder? Ive tried many ways but am not getting the correct answer.
You have to measure the area of the cylinder, that is, the outer radius area minus the inner radius.
You are given the mass, you can look up the density.
Now, solve for (radiusouter^2-radiuinner^2)
Areacopper= PI (radiusouter^2-radiusinner^2) or
So, did the size of the cylinder affect anything?
There is no inner or outer im guessing it is assumed to be negligibly thick
Nvm Thank you for the insight into the density. I got the anwser now and your right the radius has nothing to do with it.