Recent content by clive

I have to expand a plane wave in spherical waves by using e^{i\vec{k}\cdot\vec{r}}=\sum_{n=0}^{\infty}i^n (2n+1)P_n(cos \theta) j_n(kr) I wrote a MATLAB code (pasted below) in order to check this by plotting the two sides of the eq above. The graphs are similar but different. I do not know...

Just replace any occurrence of '7' in p with ' ' before disp(p): for i=1:numel(p) if(p(i)=='7') p(i)=' '; end

In the numerical approach you do not need those elliptic integrals. You just consider each loop as a collection of small tail-to-tip oriented segments and sum the individual contributions at each point (of interest) of the space: H_i=\frac{I\,\vec{dl}_i\times \hat{r}_i}{4\pi r_i^2}

I think you should give more details about your specific problem but anyway, you have to look for internal BC in your software. If this is not possible, just replace the disk with a donut whose internal radius is very small. In this way you will be able to apply regular BCs near the center of...

Here is an analytical solution: http://www.netdenizen.com/emagnet/offaxis/iloopoffaxis.htm If not, do it numerically...(and you do not need those elliptic integrals any more)

Function: double gsl_matrix_get (const gsl_matrix * m, size_t i, size_t j) This function returns the (i,j)-th element of a matrix m. If i or j lie outside the allowed range of 0 to n1-1 and 0 to n2-1 then the error handler is invoked and 0 is returned. Function: void gsl_matrix_set...

Nevermind. Just figured out: \makeatletter \renewcommand \@biblabel[1]{\bf{#1}.} \makeatother

Hi all, I have to change the bibliography style in an "article"-type tex document from [1] Author1, Author2, etc... [2] Author3, Author4, etc. ... to 1. Author1, Author2, etc. 2. Author3, Author4, etc. ... and I want to create my own .bst file using "latex makebst" command...

If "mcd" means milicandela then a "conversion" between these two quantities is impossible. I think you have to compute the luminance E (in cd/m^2) at a given point on a surface from a source with a given luminous intensity I (in cd). If this is the case, the equation you need is...

The energy emitted by the source in dt is Pdt where P=500\,kW. At R=25\,km from the source this energy will be uniformly distributed in a spherical shell of radii R and R+dR. The volume of this shell equals dV=\pi R^2 c dt where c is the speed of light. Now you have to know that the energy...

You can "feel" what's happen with the following 1-D example: \frac{d^2 V}{dx^2}=0 in [a,b] and V(a)=V(b)=V_0. (the potential between two infinite metallic plates). From the first Eq. you have V(x)=mx+n and after imposing the boundary conditions ma+n=V_0 mb+n=V_0 you get m=0 and...

If the Astronuc's demonstration didn't satisfy you, then try to solve the Laplacian equation \bigtriangledown^2 V=0 with V=V_0 on the boundary (using Femlab for example). This Eq. is obviously derived from Maxwell Eqs... The "counterexample" V(r)=r you proposed is wrong because it does...

For a very small portion of the ring (dl=Rd \theta) you have the static equilibrium condition: 2 T sin(\frac{d \theta}{2})=N where N is the force between the ring and the cylinder. Now you use sin (\frac{d \theta}{2}) ~= \frac{d \theta}{2} and T from the Young law for the small element dl...

The tension you need is F (in my eqs). F_f (the frictional force) and F (the tension in the rope) are different. You have to find F from the eq. below: \frac{F-F_f}{F_f}=\frac{m_{sled}}{m_{box}} and you'll obtain the tension in the rope.

Hi M&M The acceleration is given by the second law a=\frac{F}{m}. You can divide the F(x) graph by m and you obtain a(x). As for velocity you can apply v=\sqrt{2ax} (Galilei) until x=12 m. Beyond this limit the acceleration is decreasing linearly so you can use the same Galilei's law...