Hot answers tagged dipole
11
The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is:
\begin{equation}
U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...
6
Dipole moment is a vector and can be calculated using formula
$$\vec{p} = \sum_i q_i \vec{r}_i.$$
It can be shown easily using the formula above that in case of two charges separated by distance $d$
$$\vec{p} = q \vec{d},$$
where vector $\vec{d}$ goes starts at negative ends at positive charge.
...
6
The magnetic analog of Gauss's law tells you that
$$ \oint B dA = 0$$
This says that he number of magnetic field lines entering and leaving any surface surrounding any configuration of magnets are always equal. So there is no configuration of equal and opposite poles which produces a monopolar field. Your configuration would neutralize the magnetic field ...
5
I'll give you the derivation from my book which includes a nice way to see how the delta functions arise:
..............................................................................................................................................................
We can derive the potential field $\vec{A}$ and the electromagnetic fields $\vec{E}$ and ...
3
The potential energy in this case should be $U=+\vec{m}.\vec{B}$, hence the potential energy is minimized, as it should be. Here is the explanation:
Let’s look at the derivation of interaction energy between magnetic dipole and magnetic field carefully. The dipole energy $U=-\vec{m}.\vec{B}$ is derived using principle of virtual work with an assumption ...
3
Classically a non-pointlike spinning charged object possesses a magnetic dipole moment due to the fact that charged particles in the object are spinning around some axis. In contrast, the electron has a dipole moment that arises from its intrinsic spin angular momentum. As you point out, the electron has no internal structure, so the spin does not refer to ...
3
Because the black area is half the box below.
To explain: move the dipole from an area of no field to an area of field strength E. As you do, there's a force proportional to the dipole moment and to the gradient of E. For a fixed dipole, this force depends only on the gradient (horizontal dashed line). But for an induced dipole, the dipole moment depends ...
3
If the magnetic dipoles in a material are ordered, the material has a lower entropy because there are many fewer ways how the spins may be oriented if most of them (or all of them!) are required to be aligned.
Such an alignment also reduces the heat capacity because before the dipoles got aligned, the orientation (direction) of each dipole was a degree of ...
2
There are two misconceptions present in your explanation of the problem.
$N$ is not number of dipoles, but their volumetric density
$Q$ is not total charge, but equivalent charge at boundaries of the dielectric.
The idea is that (a) dielectric of the area $A$ and height $L$ polarized homogeneously along its height and (b) two plan-parallel plates of the ...
2
The position dependence comes in via the $R$ in the following expression¹:
$$ \vec E = \frac{3 \vec p \cdot \vec R}{4 \pi \varepsilon_0 R^3} \vec R - \frac{\vec p}{4 \pi \varepsilon_0 R^3} $$
This makes sense, as the charge $q$ also does not depend on distance - it’s effect on the electric field $\vec E$, however, does depend on said distance and the ...
2
The force on a dipole placed in an electrical field is given by $\mathbf{F} = (\mathbf{p}\cdot \nabla)\mathbf{E}$ (see, e.g., Griffiths, 3rd edition, eq. 4.5). Recall that,
$$
\nabla(\mathbf{p}\cdot\mathbf{E}) = \mathbf{p}\times (\nabla\times \mathbf{E}) + \mathbf{E}\times(\nabla\times \mathbf{p})+(\mathbf{p}\cdot\nabla)\mathbf{E} + ...
2
The dipole transition matrix element has a classical interpretation as the time Fourier series of the classical dipole moment of the Bohr orbit corresponding to one of the energy levels. The interpretation is only exact at high levels, at the correspondence limit, and the m,n matrix element is the m-n-th Fourier series coefficient for either orbit m or orbit ...
2
One must distinguish two conditions: whether the eigenvalue of $|\vec J|^2$, the squared total angular momentum, is changing; and whether the whole vector $\vec J$ is changing.
The latter is guaranteed in a dipole transition: one can't keep the whole vector constant. At most, you may satisfy the former condition: the length of $\vec J$ may stay constant so ...
2
The dipole moment of a system of charges $q_i$ located at positions $\mathbf r_i$ is defined as the vector
$$\mathbf d=\sum_i q_i\mathbf r_i.$$
If you have a single charge $q$ at $\mathbf r=d\hat{\mathbf e}$ then $\mathbf{d}$ has magnitude $qd$ and points along the unit vector $\hat{\mathbf e}$. Usually, however, this is introduced for two charges of equal ...
2
The $Q_x$ and $Q_y$ transitions are electronic excitations in the conjugated $\pi$ orbitals of the Bchl a molecule. They involve two different sets of conjugated bonds. The $Q_x$ involves a shorter chain of conjugated bonds so it occurs at a higher energy/frequency. I couldn't find a really good diagram to show which bonds are involved in the in the $Q_x$ ...
1
Equation (2.5) expresses the velocity field in function of the stream function. It's not clear to me it really should be presented at this stage in the process, I guess it's useful to impose the conditions at infinity.
Equation (2.6) expresses that the two pieces of the total solution, the one inside the disk $\psi$ and the one outside $\psi_1$, have to ...
1
Actually, if you ask about radial dependace of time-avaraged Poynting flux vector, then it depends on where the observer is (far field or near field). If you assume that the observer is far away (more than $\frac{2D^2}{\lambda}$, where $D$ - is the maximum size of your source and $\lambda$ - is the walength (I see you use monocromatic case)) than the ...
1
When introducing the stream function, the steps that you usually take are as follows.
Replace $u$ and $v$ by the streamfunction.
Derive the horizontal momentum equation (for $u$) with respect to $y$ and the other with respect to $x$.
Eliminate the pressure term, to end op with a single equation in $\psi$.
1
Nope. The electron dipole moment is defined as its intrinsic property – a property that may be imagine as a consequence of the particle's internal structure or, more generally, "something that happens inside it". So you may imagine that the relevant wave function for the "center of mass" is $\delta^{(3)}(\vec r)$. It's not literal because one can't localize ...
1
This is the old puzzle of the electromagnetic arrow of time, and it is resolved in quantum mechanics, when quantizing the electromagnetic field. The classical field starts out in a non-radiative state for the problem of dipole emission, and this is thermodyanmically infinitely improbable (what's the chance that there is no radiation at infinity?) and so the ...
1
I had learnt that the dipole moment is defined for 2 point charges only with equal magnitudes but opposite character.
Actually, that is not the case. You can calculate a dipole moment for any charge distribution.
In fact, the dipole moment is just one of a whole series of multipole moments which can be used to describe a charge distribution.
The ...
1
I’m not an expert in electromagnetism, but if one of my students (general physics, BSc level) asked, I'd say the following: the rotating proton indeed generates an magnetic field, and it behaves like a magnetic dipole for these purposes (and at large enough distances). However, that only concerns the field created by this proton, and its interaction of this ...
1
The force applied to a point dipole with dipole momentum $\vec{p}$ is
$$
\vec{F} = (\vec{p} \cdot \vec\nabla) \vec{E}
$$
In Cartesian coordinates that is
$$
F_i = \sum_j p_j \frac{\partial}{\partial x_j} E_i
$$
But in spherical coordinates it is not the same.
There is no field components along $\vec{\theta}$, but there is a gradient of field components ...
1
There are two common types of dipole vector fields in physics:
The "divergence-free" dipole field:
$\vec{V}(\vec{r}) \propto \frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi r^3} + \frac23\vec{\mu}\,\delta(\vec{r})$
The "curl-free" dipole field:
$\vec{V}(\vec{r}) \propto \frac{3\left(\vec{\mu}\cdot\hat{r}\right)\hat{r}-\vec{\mu}}{4\pi ...
1
The origin of the problem is the special point $\mathbf{r}=0$. In the usual derivation (when using easily-derived formula for the potential $\varphi = \frac{1}{4 \pi \epsilon_0} \frac{\mathbf{p} \cdot \mathbf{r}}{r^3}$):
$$E_{\alpha} = -\frac{1}{4 \pi \epsilon_0} \nabla_{\alpha} \left( \frac{p_{\beta} x_{\beta}}{r^3}\right) = -\frac{1}{4 \pi \epsilon_0} ...
1
A dipole is formed of two opposite charges. By bringing them ever closer together, all the while increasing their charge but keeping the product of the charge and the separation $p = q \times d$ constant, we can form an ideal, elementary dipole.
The field on a line in between the charges goes as $~ 1/d^4$ and thus goes to infinity as $d \to 0$. This forms ...
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