Question

A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 5.75 cm and inner radius Rb = 4.95 cm .The central conductor and the conducting tube carry equal currents of I = 3.75 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 = 5.45 cm from the axis of the conducting tube?

What is the value of the magnetic field at a distance r = 5.45 cm from the axis of the conducting tube? Recall that μ0=4π×10−7 T⋅m/A.

Express your answer numerically in teslas.

Answer 1

Let us find the total current enclosed by r = 5.45 cm
Current 3.75 A in central conductor incompletely enclosed

Amount of current enclosed in tube:
I' = pi*(5.45^2 - 4.95^2) *3.75 / {pi*(5.75^2 - 4.95^2)}
=(5.45^2 - 4.95^2) *3.75 / {(5.75^2 - 4.95^2)}
= 5.2*3.75/8.56
= 2.28 A

This is in opposite direction.

I enclosed = 3.75 - 2.28 A = 1.47 A

Use Ampere's law:
B*2*pi*r = miuo*I enclosed
B*(2*pi*5.45*10^-2) = 4*pi*10^-7 * 1.47
B*(2*5.45*10^-2) = 4*10^-7 * 1.47
B = 5.4*10^-6 T
Answer: 5.4*10^-6 T