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 R_a =6.15 cm and inner radius R_b = 5.05 cm. The central conductor and the conducting tube carry equal currents of I = 1.35 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 = 5.79 cm from the axis of the conducting tube? Correct The magnetic field lines are concentric circles around the axis of the current carrying conductors. For this reason the choice of a circular integration path dl rightarrow that is coaxial with the conductors will greatly simply your calculations. What is the value of the magnetic field at a distance r = 5.79 cm from the axis of the conducting tube? Recall that mu_0 = 4pi Times 10^-7 T m/A. Express your answer numerically in teslas. This question will

Explanation / Answer

the difference between the area pf outer and inner cylinder gives the area enclosed by the cylinder. thus, the area of the enclosing cylinder is

router2- rinner2 = (0.0615 m2- 0.0505 m2) = 3.870 * 10-3 m2

the area of the portion on of that enclosing cylinder within 5.79cm is calculated as

5.79 cm = (0.0579 m2 - 0.0505 m2) = 2.520* 10-3 m2

Therefore the portion of the cylinder within 5.79 cm carries (2.520/3.870) * 1.35 = 0.8791 A of current. Since it is in the opposite direction to that of the inner wire, the net enclosed current is 0.4709 A.

B = 0I/2r = (4 * 10-7)( 0.4709)/(2 * 0.0579) = 1.63 * 10-6 T

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