- #1
amwil
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- Homework Statement
- A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 6.85 cm and inner radius Rb = 3.75 cm . (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.85 A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 4.74 cm from the axis of the conducting tube?
What is the expression for the current Iencl enclosed in the path of integration in terms of the current I , the outer radius Ra , the inner radius Rb , and the distance from the axis r where Ra>r>Rb ?
Express your answer in terms of I , Ra , Rb , and r .
- Relevant Equations
- I = JA
I know that Ienl for the inner cylinder is just I and the current density for the outer tube is J1= -I/(pi(Ra^2-Rb^2). I assume that the current through the enclosed portion of the conducting tube (I1) is equal to J1(A1) where A1 is the area of the enclosed portion of the conducting tube. I found A1=pi(Ra^2-r^2) then multiplied it by J1 to get I1 = (-I(Ra^2-r^2))/(Ra^2-Rb^2). Then I added Iencl for the inner cylinder and I1 for the outer tube to get Iencl= I + (-I(Ra^2-r^2))/(Ra^2-Rb^2) but it said this was wrong. Can someone tell me what I'm doing wrong??