Recent content by ks_wann

Well, ##d\vec l ## and ##\hat r## are perpendicular at all times. I see that for a full circle, all the horizontal components will cancel, thus giving me ##B(z)##. Am I missing something? I'm sadly not very good at identifying and exploiting symmetries..

Okay. So I basically just state that the field points downwards in the case of z=0, and get \begin{equation} \frac{\mu_0I}{4\pi} \int_{\theta_{1}}^{\theta^{2}} \frac{R}{(R^2+z^2)^{3/2}} \sin{\theta} d \theta \end{equation}? What about the change in direction with increasing z?

So, I've got a wire with a semicircle about the origin, my task is to find the magnetic field at M=(0,0,z). I want to use Biot-Savart, so I find dl', r and r-hat. (r being the vector from the source point to field point) \begin{equation} r^{2}=R^{2}+z^{2}\\...

I did the integration in x, y and z components, and I also get 0. Is it much harder to solve the problem in cylindrical coordinates, given that "r-hat" is a function of ø? I would include the "r-hat" into the integral in my final step, find the expression for "r-hat" of ø (if one such exists...

So, I've got a charge distribution given by: \begin{equation} \rho(r,\phi,z)=\frac{q}{2\pi R}\delta(r-R)\delta(z)\cos(2\phi) \end{equation} This, if I'm not mistaken, translates into a circular charge distribution located in the z-plane, a distance R from origo. Thus \begin{equation}...

Thanks for all of your answers, it really helped me out. I've reread the series chapter of my calculus book, and I've come to a much better understanding of that subject in general. I basically expand the logarithm in the function, and then I expand the function, which gave the correct answer.

I'm a bit frustrated at the moment, as this minor problem should be fairly easy. But I seem to go wrong at some point... So I've got to do a 1st order expansion of the function \begin{equation} f=\frac{\cos(\theta)}{\sin(\theta)}\ln(\frac{L\sin(\theta)}{d\cos( \theta)}+1) \end{equation}...

Well, yes that was fairly easy. Thanks a lot. :)

I see what you did there. So I should treat \phi(\theta) properly in my derivatives.

No, I can't figure out the last term on the right.

With \theta as the parameter, I insert x=r\sin\theta\cos\phi, y=r\sin\theta\sin\phi and z=r\cos\theta. This would give me \sqrt{(\frac{d}{d\theta}r\sin\theta\cos\phi)^2+(\frac{d}{d\theta}r\sin\theta\sin\phi)^2+(\frac{d}{d\theta}r\cos\theta)^2} From there I end up with...

I've looked it up, but it puzzles me how I go from ds^2=dx^2+dy^2+dz^2 when dx, dy and dz are added together. I mean, if it was ds^2=dx^2 dy^2 dz^2 it would be straight forward for me. abs[∂(x,y,z)/∂(r,θ,phi)] * dr dθ dphi Can I simply transform each on its own? dx, dy and dz?

I'd use s=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2} So I tried inserting x, y and z as spherical coordinates, and having θ as the parameter, in the end I end up with \sqrt{R^2}. That's after I take the derivative with respect to θ. I'm not sure I got your hint.

So, I'm to show that in spherical coordinates, the length of a given path on a sphere of radius R is given by: L= R\int_{\theta_1}^{\theta_2} \sqrt{1+\sin^2(\theta) \phi'^2(\theta)}d\theta, where it is assumed \phi(\theta), and start coordinates are (\theta_1,\phi_1) and (\theta_2, \phi_2)...