Electrodynamics and vector calculus question

[goodnews]

Homework Statement

1) The magnetic field everywhere is tangential to the magnetic field lines, [itex]\vec{B}[/itex]=B[[itex]\hat{e}t[/itex]], where [[itex]\hat{e}[/itex]][/t] is the tangential unit vector. We know [itex]\frac{d\hat{e}t}{ds}[/itex]=(1/ρ)[[itex]\hat{e}[/itex]][/n]
, where ρ is the radius of curvature, s is the distance measured along a field line and [[itex]\hat{e}[/itex]][/n] is the normal unit vector to the field line.

Show the radius of curvature at any point on a magnetic field line is given by ρ=[itex]\frac{B^3}{abs(\vec{B}X(\vec{B}\bullet\vec{B})\vec{B}) }[/itex]

Homework Equations

[itex]\vec{B}[/itex]=B[[itex]\hat{e}[/itex]][/t]
[itex]\frac{d\hat{e}t}{ds}[/itex]=(1/ρ)[[itex]\hat{e}[/itex]][/n]
ρ=[itex]\frac{B^3}{abs(\vec{B}X(\vec{B}\bullet\vec{B})\vec{B}) }[/itex]

The Attempt at a Solution

solved the vector equation, and would then use some form of stokes theorem to equate it and find the value of ρ

 

[lippyka]

. We can write the vector equation as B^2[\hat{e}][/t]=\frac{d}{ds}(B[\hat{e}][/t])=\frac{1}{ρ}[\hat{e}][/n]Now using stokes theorem we get ∮(B[\hat{e}][/t])Xd[\vec{s}]=∮[\hat{e}][/n]dAwhere A is the area of the region. Now rearranging the equation we get ρ=\frac{B^2s}{A}where s is the length of the path. Now substituting the value of s in the original equation we get ρ=\frac{B^3}{abs(\vec{B}X(\vec{B}\bullet\vec{B})\vec{B}) }