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 = 6.65 cm and inner radius Rb = 5.05 cm . (Figure 1) The central conductor and the conducting tube carry equal currents of I = 2.55 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 = 5.35 cm from the axis of the conducting tube?

What is the value of the magnetic field at a distance r = 5.35 cm from the axis of the conducting tube? Recall that ?0=4?×10?7 T?m/A.

Explanation / Answer

Imagine a amperian loop with radius r = 5.35 cm
The current enclosed by the the loop,

I_enclosed = I - I*pi*(r^2 - Rb^2)/(pi*(Ra^2 - Rb^2))

= 2.55 - 2.55*(0.0535^2 - 0.0505^2)/(0.0665^2 - 0.0505^2)

= 2.125 A

now use Ampere's law

integral B.dL = mue*I_enclosed

B*2*pi*r = mue*I_enclosed

B = mue*I_enclosed/(2*pi*r)

= 4*pi*10^-7*2.125/(2*pi*0.0535)

= 7.94*10^-6 T

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