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  1. R

    Derivation: force on magnetic moment in magnetic field

    So here it goes. First, some required vector identities: The Jabobi identity: $$\vec{r} \times (d\vec{l} \times \vec{B}) + d\vec{l} \times (\vec{B} \times \vec{r}) + \vec{B} \times (\vec{r} \times d\vec{l}) = \vec{0},$$ and the differential: $$d[\vec{r} \times (\vec{r} \times \vec{B})] =...
  2. R

    Derivation: force on magnetic moment in magnetic field

    Right, I agree completely. I forgot about my undergraduate electrodynamics book, we used Griffiths hehe. In Griffiths, the derivation is done for a rectangular loop which is sloped with respect to the magnetic field. However, problem 2 from the chapter 'Magnetic Fields in Matter' covers this...
  3. R

    Derivation: force on magnetic moment in magnetic field

    I am trying to derive the equation ##\vec{\Gamma} = \vec{m} \times \vec{B}##, where ##\vec{m} = I \vec{A}## is the magnetic moment, and ##\vec{A}## is normal to surface ##A##, from the Lorentz force law ##\mathrm{d}\vec{F} = I \vec{dl} \times \vec{B}##. For an arbitrary closed current loop ##C##...
  4. R

    A Asymptotic solutions to a differential equation

    I have been able to find the solution. Will try to post the solution if time allows for those interested. Thank you nevertheless.
  5. R

    A Asymptotic solutions to a differential equation

    I am asked to solve the differential equation $$ f''(\eta)+\frac{f'(\eta)}{\eta}+\Big(1-\frac{s^2}{\eta^2}\Big) f(\eta) - f(\eta)^3 = 0, $$ for small ##\eta## and large ##\eta## under the condition ##f(\eta \rightarrow \infty) = 1## and ##f(0)=0##. The numerically solved solution looks like...
  6. R

    Diffusion equation in polar coordinates

    Written in Cartensian coordinates, the vorticity diffusion equation for ##\omega(x,y,t)## reads $$\frac{\partial \omega}{\partial t} = \nu \Big( \frac{\partial^2 \omega}{\partial x^2} + \frac{\partial^2 \omega}{\partial y^2} \Big).$$ Fourier transforming the ##x,y## coordinates to ##k_x,k_y##...
  7. R

    Diffusion equation in polar coordinates

    Thanks for the replies. Using Orodruin's approach of doing it in Cartesian coordinates separates the problem nicely into a problem I can solve and is a lot more intuitive than using polar coordinates.
  8. R

    Diffusion equation in polar coordinates

    Thanks, that makes sense. What would be the easiest way to solve the problem under the initial condition ##\omega(t=0)=\gamma \delta(\vec{r})##?
  9. R

    Diffusion equation in polar coordinates

    Homework Statement I am trying to solve the axisymmetric diffusion equation for vorticity by Fourier transformation. Homework Equations $$ \frac{\partial \omega}{\partial t} = \nu \Big( \frac{1}{r}\frac{\partial \omega}{\partial r} + \frac{\partial^2 \omega}{\partial r^2} \Big). $$ The...
  10. R

    Taylor expansion

    Ah, I understand. Wish I had more practise with these expansion. Thanks for the help all!
  11. R

    Taylor expansion

    Ok, so that yields me $$f(x) = \frac{x^4}{x^2+x^4/12},$$ now what? ;-)
  12. R

    Taylor expansion

    Thanks for the help mfb!I give up... I am studying for a thermal physics test, and this really does not contribute to my understanding anymore. ;-)
  13. R

    Taylor expansion

    Clever Ray! Let's see whether I am getting it. $$x^4 \frac{1+x+x^2/2}{(x+x^2/2+x^3/6)^2} = x^2 (1+x+x^2/2) (1+x/2+x^2/6)^{-2} = x^2 (1+x+x^2/2) (1-2(x/2+x^2/6)) = x^2 (1+x+x^2/2) (1-x-x^2/3) = (x^2+x^3+x^4/2)(1-x-x^2/3)=x^2-x^3-x^4/3+x^3-x^4+x^4/2=x^2-10/12 x^4. $$ Nearly there, did I miss...
  14. R

    Taylor expansion

    Yes, thank you mfb. I tried: $$x^4 \frac{1+x+O(x^2)}{(x+O(x^2))^2} = x^2 + x^3,$$ which is clearly not the way to go. My main question is how to deal with that the expansion of ## e^x-1 ## will be squared and how that influences my choice of developing up till certain order terms. Can this...
  15. R

    Taylor expansion

    Hello friends, I need to compute the taylor expansion of $$\frac{x^4 e^x}{(e^x-1)^2}, $$ for ##x<<1##, to find $$ x^2 + \frac{x^4}{12}.$$ Can someone explain this to me? Thanks!
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