A very long coaxial cable consists of a solid cylindrical inner conductor of radius 3.8mm ,...

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Physics

A very long coaxial cable consists of a solid cylindrical innerconductor of radius 3.8mm , surrounded by an outer cylindricalconductor with inner radius 7mm , and outer radius 9.3mm . Theregion between the two conductors is filled with a waxlikeinsulating material to keep the conductors from touching eachother. The inner conductor carries a current 4.65A , into the page,while the outer conductor carries a current 4.6A , out of thepage.

Part A

Use Amp�re's law to derive an expression for the magnetic fieldas a function of r(the distance from the central axis) atpoints between the two conductors(R1<r<R2i). Use thisexpression to find the magnetic field at 4.8mm .

B =

1.94�10^(-4)

T [I GOT THIS PARTCORRECT]

Use Amp�re's law to derive an expression for the magnetic fieldas a function of r(the distance from the central axis) atpoints inside the outer conductor(R2i<r<R2o). Usethis expression to find the magnetic field at 7.1mm .

Use into page as negative and out of page as positivecurrent.

B =T

Comment

Hint 1. Ampere's Law

Ampere's Lawstates:  ?B??s=?0?IenclosedTherefore you need to figure out what the enclosed current is forthis case.

Hint 2. Enclosed current

The enclosed current is the current in the inner conductor +that portion of the current enclosed in the outer conductor. UseItotal/Atotal=Ienclosed/AenclosedDon't forgetAtotal=pi?(R22o?R22i)and there is a corresponding expression forAenclosed.

Answer & Explanation Solved by verified expert
3.6 Ratings (272 Votes)
for outer cylindrical conductor Ri 7 mm 7 x 103 mRo 93 mm 93 x 103 mArea of crosssection of outer cynlindrical conductor is givenas Aouter Ri2 R2oAouter 314 93 x    See Answer
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