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Chapter 4: Q43E (page 950)

Coaxial Cable. A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c (Fig. E28.43). The central conductor and tube carry equal currents I in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Derive an expression for the magnitude of the magnetic field (a) at points outside the central, solid conductor but inside the tube a鈭�r鈭�band (b) at points outside the tube.r鈮�c

Short Answer

Expert verified

A) The EXPRESSION FOR THE MAGNITUDE OF THE MAGNETIC FIELD IS PROVED

Step by step solution

01

Concepts of Ampere's law

According to this law the circulation fB鈫�dl鈫�渭0iof a resultant magnetic field along a close loop is equal totimes the total current. Here 渭0 is the permeability of vacuum, i is total current and B is, magnetic field.

02

EXPRESSION FOR THE MAGNITUDE OF THE MAGNETIC FIELD:-

Case 1 Let the magnetic field is B ata鈭�r鈭�b so now we can apply Ampere's law, we get

鈭�B鈫�dl=B(2蟺r)鈭�B鈫�dl=渭0lenclosedfB=渭0lenclosedB(2蟺r)=渭enclosed(2蟺r)

Therefore, the final result is B=渭0l(2蟺r)

Case 2 Let the magnetic field is B at c鈭�rif we are going counter clock wise, then the direction of current is out of page,

Ienclosed=l鈭�I=0fB鈫�dl鈫�0which meansB=0

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Fig. E25.29.
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