± PSS 28.2 Ampere\'s Law Submit Give Up Learning Goal: Correct To practice Probl
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Question
± PSS 28.2 Ampere's Law Submit Give Up Learning Goal: Correct To practice Problem Solving Strategy 28 2 Amperes Law. 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 that is coaxial with the conductors will greatly simply your calculationis A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra 6.15 cm and inner radius Ry-455 cn Figure 1) The central conductor and the conducting tube carry equal currents of I 405 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- 4.95 em from the axis of the conducting tube? EXECUTE the solution as follows Part c What is the value of the magnetic field at a distance r- 4 95 em from the axis of the conducting tube? Recal that-4MX 10-7 T-m/A Express your answer numerically in teslas Hints Eoure 1lli o B- Submit My Answers Sve lUp Incorrect: Try Again; 15 attempts remaining EVALUATE your answer Part D This queston w·be shown after you complete previous question(s) Proide Feedback CExplanation / Answer
c)
Area enclosed by the cylinder
A1=pi*0.06152-pi*0.04552=5.3784*10-3 m2
Area of enclosing cylinder
A2=pi*0.04952-pi*0.04552=1.1938*10-3 m2
Current carried by enclosing cy'inder
I2=(1.1938/5.3784)*4.05=0.9 A
Net enclosed current
Ien=4.05-0.9=3.15 A
Therefore magnetic field strength is
B=uoI/2pir =(4pi*10-7)*3.15/2pi*0.0495
B=1.273*10-5T
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