# Tag Info

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The relationship you propose only applies if the current has symmetry with respect to the axis of the wire. Amperes law is that the line integral of the B-field around a closed loop equals the enclosed current times $\mu$. $$\oint \vec{B}\cdot d\vec{l} = \mu I$$ To get your relationship you need to assume that on any circular path around the axis, that ...

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There can be in principle both magnetic sources and electric sources. Including these, one may write the full Maxwell equations as $d \star F = \star j_E,$ $d F = \star j_M,$ where $j_{E,M}$ are the electric/magnetic currents, respectively (as I've written them they are 1-forms). Since there are no experimentally observed magnetic charges (magnetic ...

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If one particle is fixed, some force is keeping it fixed, and in the presence of an external force, conservation of momentum doesn't apply. Your second equation is then $v_1 = 0$ (assuming particle #1 is the one that is fixed), not $m_1 v_1 + m_2 v_2 = 0$.

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In the first case,momentum is conserved because force is applied to each charge from WITHIN the system.So,the center of mass of the system is constant.In the second case,in order for just one charge to move,it has to be put in an EXTERNAL electric field.So,you can see that momentum within the system which consists only of one charge can not be conserved.If ...

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The incorrect part of your statements is that every current element does not have a circular magnetic field around it. Only an infinitely long wire would have perfectly circular magnetic field lines around it. In fact every current element does contribute to the field at point P and has a component both perpendicular and parallel to the wire. However, for ...

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A bit of clicking gets you to http://www.ru.nl/hfml/research/levitation/diamagnetic/ which tells us that the frog was levitating in a field of 16 Tesla. They give the math as well: Therefore, the vertical field gradient ∇B2 required for levitation has to be larger than $2µ_0ρg/χ$. Molecular susceptibilities χ are typically 10$^{-5}$ for diamagnetics and ...

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My guess; you are mixing up quadripoles and quadrupoles. Quadripoles are two-port networks used in electric circuit analysis. The original German word is "Vierpol Theorie", which means Four-pol because of 4 Poles. https://en.wikipedia.org/wiki/Two-port_network Quadrupoles are related to multipole expansion used in electromagnetic, atomic orbital,.. theory. ...

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Note that a current carrying wire produces a circular magnetic field that's why it doesn't matter how you hold your hand ie how you rotate your hand around your arm as long as your thumb shows the direction of the current. Edit after comments: See the illustration I've added below. Now use your hand in the way that you've learned and convince yourself ...

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let's call the circuit in the origin circuit one and it's line element $dl_1=(0,dy_1,0)$ and the one to it's right $r_2=(d,y_2,0)$ then the force between them is $dF_{12}=i dl_2 \times B_1$ where $$B_1=\frac{\mu_0i}{4\pi}\int_{l_1} \frac{dl_1\times \Delta r}{\Delta r^3}$$ and $\Delta r=(d,(y_2-y_1),0)$ so we have that $$dl_1\times \Delta r=(0,0,-dy_1 d)$$ so ...

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The minus sign is wrong.The reason for this is the x which you have chosen to be positive but is in fact negative. x points positively to the right and negatively to the left,and the horizontal vector that you are using in your picture is opposite to the direction of x.So,its x=-acotθ. Cheers!

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You might find the Faraday Effect easier. Since it was actually done by Faraday it is quite easy. You need a glass or plastic rod aligned with the magnetic field you want to measure, a monochromatic or coherent light source, a couple of polarizing filters and a photodiode. The magnetic field rotates the plane of polarization proportional to the field in the ...

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This equation: $$m_1 v_1 + m_2 v_2 = 0$$ will not work that way. There are multiple ways of looking at a system of moving particles: You can look at it in reference to a random origin, for example one corner of your room. This is what we usually do. And this is where you are trying to apply your equation, but it doesn't work. You can set your origin to the ...

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We can very well distinguish between magnetic and electric fields by simply observing the movement of a charged particle. As it was said correctly, the magnetic field does not do any work on the particle, meaning it does not accelerate it linearly. It does however apply a force that is perpendicular to the direction of motion. To illustrate, we assume a ...

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The author confused things in section 3.1. In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane): $$I = \overline{\mathbf S \cdot \mathbf n}$$ As you have shown, this is function of $|\mathbf E|^2$ only in some ...

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My question is that if we repeat the above procedure in vacuum, would the sparks be produced? Not in perfect vacuum, because there is no gas to ionize. An understanding of this behavior can be attained by studying Paschen's Law (see http://en.wikipedia.org/wiki/Paschen%27s_law), which describes the breakdown voltage for given distance and pressure.

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You have forgotten that the vector $\vec{{S}}$ is actually a function of $\phi$. If you quickly switch into cartesian coordinates $\vec{S}=\cos\left(\phi\right)\vec{i}+\sin\left(\phi\right)\vec{j}$ you can see that when you integrate the $\vec{S}$ component around $2\pi$ the sine and cosine give zero. You have to be quite careful when using curvelinear ...

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The dielectric constant, or more appropriately, the dielectric function, can be thought of as a measure of screening. A simple relation for which to picture this is: $V_{eff} = V_{ext}/\epsilon$ Therefore, in TMDs, since the electrons are more mobile in the planes, they tend to screen potentials with a greater efficiency. This gives a higher dielectric ...

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Let us assume that the magnetic field we're talking about is homogeneous in the relevant regions of space. To create such magnetic field in a smaller box requires less energy than to create the same magnetic field in a larger (that is with a larger support as a vector field) box. This is why you clearly have more energy in the larger box than the smaller box ...

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A few points of confirmation/correction: Yes the electrons flow in a direction opposite to the "conventional" current. No there is no "deficiency" if electrons - rather they have a different "potential" which is caused by the chemical reactions in the battery. No you don't have to invoke "surface charges" in the wire in order to understand current - ...

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To make the correct answer clearer, allow me to introduce the canonical momentum $\vec{p}$, given by: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}$$ This way we can rewrite the Hamiltonian as: $$H=\vec{p}\cdot\vec{\dot{x}}-L$$ Let's start by computing $\vec{p}$: $$\vec{p}=\dfrac{\partial L}{\partial\dot{x}}=m\vec{\dot{x}}+\dfrac{e}{c}\vec{A}(\vec{x},t)$$ ...

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No it is not correct because your Amperian loop is wrong.The circle has a more difficult integral.You can not get B out of the integral in a circular Amperian loop because in each point there is a different angle between B and dl,so the dot product between them changes. By saying that it is a long thin sheet i assume that it has infinite dimensions and zero ...

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Without being sure of my answer,i would say that this difference in magnitude has to do with the charges themselves. The first time you did it,you pushed negative charges in one direction.When you reversed the polarity,then you already had a situation which was not in equilibrium as the first time.The first time you had a flow in which electrons and protons ...

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