/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A wire of radius \(R\) carries c... [FREE SOLUTION] | 91影视

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Chapter 27: Problem 67

A wire of radius \(R\) carries current \(I\) distributed uniformly over its cross section. Find an expression for the total magnetic energy per unit length within the wire.

Short Answer

Expert verified
The total magnetic energy per unit length within the wire, given a uniformly distributed current, is expressed as \(U/L = mu_0I^2/8\).

Step by step solution

01

Define the Magnetic Energy Density

The magnetic energy density \(u_B\) is given by \(u_B = B^2/(2mu_0)\), where \(B\) is the magnetic field and \(mu_0\) is the vacuum permeability. However, we need to substitute \(B\) with the magnetic field inside the wire, obtained using Ampere's Law.
02

Calculate the Magnetic Field Inside the Wire

Apply Ampere's law in the form \(B(2蟺r) = mu_0Ir\) to the wire's cross-section to find the magnetic field \(B\) inside the wire. Solving this equation gives that \(B = mu_0I(r/R^2)\), where \(r\) is less than \(R\), the radius of the wire.
03

Substitute the Magnetic Field Into the Energy Density Formula

Substitute the magnetic field inside the wire into the energy density formula to get \(u_B = (mu_0^2I^2r^2)/(2mu_0R^4)\) = \(mu_0I^2r^2/(2R^4)\).
04

Integrate the Energy Density Over the Wire's Volume

Now, we need to integrate this energy density over the volume of the wire to find the total magnetic energy per unit length within the wire. This means that we need to integrate the expression for \(u_B\) times the infinitesimal volume element in cylindrical coordinates, \(r dr d蠁 dz\), over the volume of the wire, and then divide the result by the length of the wire. So, we need to calculate \(U/L\) = \(鈭玙0^R (mu_0I^2r^2/(2R^4)) r dr\) = \(mu_0I^2/(2R^4) 鈭玙0^R r^3 dr\).
05

Evaluate the Integral

Evaluating the integral \(鈭玙0^R r^3 dr\) gives \(R^4/4\). Thus, \(U/L\) = \(mu_0I^2/(2R^4)(R^4/4) = mu_0I^2/8\).

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Most popular questions from this chapter

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