cutdown

{
 "metadata": {
  "name": "Cutdown"
 },
 "nbformat": 3,
 "nbformat_minor": 0,
 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ I = \\frac{\\epsilon}{(R+r)} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ P = I^2R $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ P = \\frac{\\epsilon^2R}{(R+r)^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\Phi A = \\frac{\\epsilon^2R}{(R+r)^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ A = \\pi dL $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\Phi \\pi dL = \\frac{\\epsilon^2R}{(R+r)^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ R = \\frac{\\rho L}{A} = \\frac{\\rho L}{0.25\\pi d^2} = \\frac{4\\rho L}{\\pi d^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\Phi \\pi dL (R+r)^2 = \\epsilon^2 R $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\Phi \\pi dL \\left(\\frac{4\\rho L}{\\pi d^2}+r\\right)^2 = \\epsilon^2 \\frac{4\\rho L}{\\pi d^2}   $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\Phi \\pi d\\left(\\frac{4\\rho L}{\\pi d^2}+r\\right)^2 = \\frac{\\epsilon^2 4\\rho}{\\pi d^2}   $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\Phi \\pi d\\left(\\frac{16\\rho^2L^2}{\\pi^2d^4} + \\frac{8\\rho Lr}{\\pi d^2} + r^2\\right) = \\frac{\\epsilon^2 4\\rho}{\\pi d^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\left(\\frac{16\\Phi \\pi d\\rho^2L^2}{\\pi^2d^4} + \\frac{8\\Phi \\pi d\\rho Lr}{\\pi d^2} + \\Phi \\pi dr^2\\right) = \\frac{\\epsilon^2 4\\rho}{\\pi d^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\left(\\frac{16\\Phi\\rho^2L^2}{\\pi d^3} + \\frac{8\\Phi\\rho Lr}{d} + \\Phi \\pi dr^2\\right) = \\frac{\\epsilon^2 4\\rho}{\\pi d^2} $$"
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": "#$$ \\frac{16\\Phi\\rho^2}{\\pi d^3}L^2 + \\frac{8\\Phi\\rho r}{d}L + \\left(\\Phi \\pi dr^2 - \\frac{\\epsilon^2 4\\rho}{\\pi d^2}\\right) = 0 $$"
    }
   ],
   "metadata": {}
  }
 ]
}