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 $a ∠ r ∠ b$ and (b) at points outside the tube. $r \geq c$

Short Answer

Expert verified

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

Step by step solution

Concepts of Ampere's law

According to this law the circulation $f B \to dl \to μ_{0} i$ of a resultant magnetic field along a close loop is equal totimes the total current. Here $μ_{0}$ is the permeability of vacuum, i is total current and B is, magnetic field.

EXPRESSION FOR THE MAGNITUDE OF THE MAGNETIC FIELD:-

Case 1 Let the magnetic field is B at $a ∠ r ∠ b$ so now we can apply Ampere's law, we get

$\begin{aligned} \int B \to d l = B (2 π r) \\ \int B \to d l = μ_{0} l_{enclosed} \\ f B = μ_{0} l_{enclosed} \\ B (2 π r) = μ_{enclosed} (2 π r) \end{aligned}$

Therefore, the final result is $B = μ_{0} l (2 π r)$

Case 2 Let the magnetic field is B at $c ∠ r$ if we are going counter clock wise, then the direction of current is out of page,

$\begin{aligned} I_{enclosed} = l - I = 0 \\ fB \to dl \to 0 which means B = 0 \end{aligned}$

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Short Answer

Step by step solution

Concepts of Ampere's law

EXPRESSION FOR THE MAGNITUDE OF THE MAGNETIC FIELD:-

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