- #1
pondzo
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Homework Statement
A dipole of moment ##\vec{p}## where p is fixed, moves with velocity ##\vec{v}## though a magnetic field ##\vec{B}##. Show that the force on the dipole is ##\vec{v}\times(\vec{p}\cdot\vec{\nabla})\vec{B}+\dot{\vec{p}}\times\vec{B}##.
Homework Equations
##\vec{F}=e(\vec{E}+\vec{v}\times\vec{B})##
The Attempt at a Solution
This is how I am picturing the situation:
##\vec{F}=\sum_i\vec{F_i}##
##~~~=e\vec{v}\times\vec{B_1}-e\vec{v}\times\vec{B_2}##
##~~~=e\vec{v}\times\vec{B}(\vec{a}+\frac{1}{2}\vec{d})-e\vec{v}\times\vec{B}(\vec{a}-\frac{1}{2}\vec{d})##, Now I taylor expand ##\vec{B}## and disregard high order terms:
##~~~\approx e\vec{v}\times[\vec{B}(\vec{a})+\frac{1}{2}\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}}]-e\vec{v}\times[\vec{B}(\vec{a})-\frac{1}{2}\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}}]##
##~~~=e\vec{v}\times\vec{B}(\vec{a})+\frac{1}{2}e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})-e\vec{v}\times\vec{B}(\vec{a})+\frac{1}{2}e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})##
##~~~=e\vec{v}\times(\vec{d}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})##
##~~~=\vec{v}\times(\vec{p}\cdot\vec{\nabla}\vec{B}|_{\vec{a}})##
Using the identity ##\vec{\nabla}\times(\phi\vec{B})=(\vec{\nabla}\phi)\times\vec{B}+\phi\vec{\nabla}\times\vec{B}## and replacing ##\vec{\nabla}## with ##\vec{v}##, also replacing ##\phi## with ##(\vec{p}\cdot\vec{\nabla})## we get:
##~~~=\vec{v}(\vec{p}\cdot\vec{\nabla})\times\vec{B}+(\vec{p}\cdot\vec{\nabla})\vec{v}\times\vec{B}##
##~~~=\vec{v}\times (\vec{p}\cdot\vec{\nabla})\vec{B}+(\vec{p}\cdot\vec{\nabla})\vec{v}\times\vec{B}##
I think the above derivation is correct so far, so all that remains is to show that ##(\vec{p}\cdot\vec{\nabla})\vec{v}\equiv \dot{\vec{p}}## But I am stuck with this, any help will be appreciated.