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...
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##...
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...
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.
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...
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!