Learning Goal: To practice Problem-Solving Strategy 28.2 Ampere\'s Law. A solid

Question

Learning Goal:

To practice Problem-Solving Strategy 28.2 Ampere's Law.

A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 6.25cm and inner radius Rb = 3.25cm .    (Figure 1) The central conductor and the conducting tube carry equal currents of I = 1.35A 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 = 4.23cm from the axis of the conducting tube?

Part C

What is the value of the magnetic field at a distance r = 4.23cm from the axis of the conducting tube? Recall that ?0=4?

Explanation / Answer

The area of the enclosing cylinder = ?router2- ?rinner2 = ?(0.0625 m2 - 0.0325 m2) = 8.953 * 10-3 m2

The area of the portion of that enclosing cylinder within 4.23 cm = ?(0.0423 m2 - 0.0325 m2) = 2.303 * 10-3 m2

Therefore the portion of the cylinder within 4.23 cm carries (2.303/8.953) * 1.35 = 0.3473 A of current. Since it is in the opposite direction to that of the inner wire, the net enclosed current is 1.35 - 0.3473 = 1.003 A.

B = ?0I/2?r = (4? * 10-7)(1.003)/(2? * 0.0423) = 4.74 * 10-6 T

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