Radius of Curvature calculation in a magnetic field

[goodnews]

Homework Statement

1) The magnetic field everywhere is tangential to the magnetic field lines, $\vec{B}$=B$\hat{e}t$, where $\hat{e}t$ is the tangential unit vector. We know $\frac{d\hat{e}t}{ds}$=(1/ρ)$\hat{e}n$
, where ρ is the radius of curvature, s is the distance measured along a field line and [$\hat{e}$][/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=$\frac{B^3}{abs(\vec{B}X(\vec{B}\bullet\vec{B})\vec{B}) }$

Homework Equations

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

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

 

[member 553083]

.From the given vector equation, we have:\vec{B}=B[\hat{e}][/t]\frac{d\hat{e}t}{ds}=(1/ρ)[\hat{e}][/n]Then, using Stokes theorem, we have:\oint\vec{B}\bullet d\vec{s} = \int\int{curvature*dA}Where dA is the area of the loop and curvature is 1/ρ.Plugging in the given values for \vec{B} and d\vec{s}, we get:B\oint\hat{e}t\bullet [\hat{e}][/t]ds = \int\int{1/ρ*dA}Simplifying this equation, we get:B*ds = \int\int{1/ρ*dA}Rearranging the equation, we get:ρ = \frac{B^3}{abs(\vec{B}X(\vec{B}\bullet\vec{B})\vec{B}) }