Difference between revisions of "Chapter 30 Problem 19"

From 105/106 Lecture Notes by OBM
 
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==Solution==
 
==Solution==
Since the energy density is a function of radius only, we treat the toroid as cylindrical shells each with differential volume <math>dV=2 \pi rhdr</math> .
 
  
 
<math>\oint \vec{B}.d\vec{l}=\mu_0 I_{\textrm{encl}}</math>
 
<math>\oint \vec{B}.d\vec{l}=\mu_0 I_{\textrm{encl}}</math>
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<math>u_b=\frac{B^2}{2\mu_0}=\frac{1}{2\mu_0}\left( \frac{\mu_0 NI}{2 \pi r}\right)^2=\frac{\mu_0 N^2 I^2 }{8\pi^2 r^2}</math>
 
<math>u_b=\frac{B^2}{2\mu_0}=\frac{1}{2\mu_0}\left( \frac{\mu_0 NI}{2 \pi r}\right)^2=\frac{\mu_0 N^2 I^2 }{8\pi^2 r^2}</math>
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Since the energy density is a function of radius only, we treat the toroid as cylindrical shells each with differential volume <math>dV=2 \pi rhdr</math> .
  
 
<math>U=\int u_b dV=\int_{r_1}^{r_2} \frac{\mu_0 N^2 I^2 }{8\pi^2 r^2} 2 \pi r h dr= \frac{\mu_0 N^2 I^2 h}{4\pi }  \int_{r_1}^{r_2}  \frac{dr}{r}=\frac{\mu_0 N^2 I^2 h}{4\pi }\ln\left(\frac{r_1}{r_2}\right)</math>
 
<math>U=\int u_b dV=\int_{r_1}^{r_2} \frac{\mu_0 N^2 I^2 }{8\pi^2 r^2} 2 \pi r h dr= \frac{\mu_0 N^2 I^2 h}{4\pi }  \int_{r_1}^{r_2}  \frac{dr}{r}=\frac{\mu_0 N^2 I^2 h}{4\pi }\ln\left(\frac{r_1}{r_2}\right)</math>

Latest revision as of 22:51, 5 May 2019

Problem

A toroid of rectangular cross section, with N turns carrying a current I.

For the toroid of Problem 13, determine the energy density in the magnetic field as a function of () and integrate this over the volume to obtain the total energy stored in the toroid, which carries a current in each of its loops.

Solution

Since the energy density is a function of radius only, we treat the toroid as cylindrical shells each with differential volume .