Recent content by UgOOgU

What you have to do is calculate B in the points of the segments AB and CD. dl is not upper these segments, is upper the wire that generate B. B = \oint_{wire}...

Unfortunatelly no. B is equal to \frac{\mu _0 I}{2\pi r_1} solely in the case of an infinitely long wire. In the present case, you have to realize the Biot-Savart integration. Try, it is not difficult. P.S.: "Usted hablas español, my amigo?"

Ok, now I understand, \widehat{\eta} is the versor of the vector \vec{\eta}. Sorry, I think that it was wrong because of your substitution in this equation that had to be \vec{E}(r)=\frac{1}{4\pi\epsilon_{0}}\int_V\frac{\rho}{x^2+y^2+z^2}dx dy dz.

To see how the unit systems of electromagnetic quantities are constructed and understand the relations between its equations I recommend: Jackson, J.D. Appendix on Units and Dimensions on Classical Electrodynamics.

No, the equation is really without it. E(r)=\frac{1}{4\pi\epsilon_{0}}\int_{V}\frac{\rho(r')}{\eta^2} d\tau'

This equation is not OK. The \eta in the upper part of the fraction do not exists.

If I had understand the problem, what you are looking for is calculate the integral \oint_C \vec{B}\cdot\vec{dl}. CD and AB are not orthogonal to B. Your intuition is wrong. B = \frac{\mu_0 I}{2\pi r_1} and B = \frac{\mu_0 I}{2\pi r_2} in the parts AB and CD respectively. B is not pointing...

What you are looking for is the amplitude, not the energy. What you have said about energy is right, now you have to use the relation between energy and amplitude, do you remember it?

Yes, photons are massles. But the more fundamental concept in applying forces is not mass or inertia; on the contrary, is the momentum, because force changes the momentum \vec{F} = \frac{d\vec{p}}{dt}. Although photons are massless, they have momentum \vec{p} = \hbar\vec{k}, the necessary...

How? In the manner that I have defined in my prior post. In there I had established the cross products of the base of the space, it has not reference with hands or with the "common representation" of these base.

I think that the right-hand rule is a consequence of the definition of the coordinate system. The three-dimensional euclidian vector space that is usually used in physics is by definition a "right-handed coordinate system". In other words, the versor products of the base 'i x j = k' , 'j x k =...

You have an integral whose integrand is |f(x)|^{2}. If f(x) = 0 everywhere, the integral is obviously zero. If f(x) \neq 0 in at least one point, the integrand is |f(x)|^{2}>0 in at least that point, being zero in the others. Your interval of integration is positive, dx > 0 everywhere. So...

No, if you remember the definition of the integral as a limit of a riemannian sum, you will understand what aPhilosopher is trying to say.

The distance between two points is the magnitude of the vector connecting these two points. In your problem this vector is \vec{r} = \vec{r_{2}}-\vec{r_{1}}. To encounter the magnitude of this vector, all we have to do is calcute the scalar product of it with itself and take the square root. So...