A solid cylindrical conductor is supported by insulating disks on the axis of a

Question

A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 7.55cm and inner radius Rb = 3.95cm . (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.15A 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 = 5.90cm from the axis of the conducting tube?

A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 7.55cm and inner radius Rb = 3.95cm . (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.15A 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 = 5.90cm from the axis of the conducting tube?

Explanation / Answer

Current density(J) is the current(I) per area(A).
So J=I/A, however your A is the area of the circle. So we use the equation pi*r^2=A
J=I/pi(r^2)
Its in cm so I will change it just by dividing by a hundred.

1) The inner wire has a radius of 5.9/100, so J=2.15/(pi*(5.9/100)^2 =196.60

2) For the outer shell is J=2.15/(pi*[(3.95/100)^2-(7.55/100)^2]) = -2.79*10^-2

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