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 = 7.25cm 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.

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.

Explanation / Answer

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]

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