A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 7.25cm and inner radius Rb = 4.55cm The central conductor and the conducting tube carry equal currents of I = 1.45A 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 = 5.29cm from the axis of the conducting tube?
What is the value of the magnetic field at a distance r = 5.29cm from the axis of the conducting tube? Recall that ?0=4?�10?7 T?m/A.
Express your answer numerically in teslas.
Note that
B = [uo I] / [2 pi R]
where I is the current enclosed by the Gaussian cylinder with
radius R = 5.29 cm.
We know it encloses the whole of the inner conductor.
However, this only encloses some part of the outer
conductor.
Here, the cross section area of the outer conductor is A = pi(Rb^2
- Ra^2). Thus, the total cross section area of the outer cylinder
is
A = 100.0911419
cm^2
Also, the area encloses is A_enc = pi(R^2 -Ra^2). Thus,
A_enc = 22.87582107
cm^2
Thus, the fractional part enclosed is
%enclosed = 0.228549906
Thus, the current of the outer conductor enclosed is
only
I_enc = 1.45(0.22855) =
0.331397363 A
Thus, the net current enclosed is, as the inner and outer
conductors have opposite currents, is
I_enctotal = 1.45 - 0.3314 A =
1.118602637 A
Thus, the magnetic field is
B = [uo I]/[2 pi R] = 4.23E-06
T [ANSWER]
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 mm so I will change it just by dividing by a thousand.
1) The inner wire has a radius of 0.725/1000, so
J=5.4/(pi*(0.725/1000)^2
2) For the outer shell is
J=1.9/(pi*[(9.9/1000)^2-(9.1/1000)^2])
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