Let $L , I , \phi$ are self inductance , current and magnetic flux (created by each loop) of a solenoid with $n$ turns . $r$ is the radius of those loops . We have , $LI=n \phi = nAB $  [$A=\pi r^2$]  Now according to textbook , "As there are n loops ,  $\displaystyle B=\frac{n\mu_oI}{2r}$ .   Substituting $B$ , $\displaystyle LI=nA\times\frac{n\mu_oI}{2r}\Rightarrow L=\frac{n^2\mu_o \pi r}{2}$ "  We considered the fact of n loops when we multiplied $n$ with $\phi$ . Again we multiplied $n$ while calculating $B$ .   In the equation $n\phi=nAB$ , $B$ represents magnetic flux density caused by each loop . On the other hand , in the equation $\displaystyle B=\frac{n\mu_oI}{2r}$ , $B$ represents magnetic flux density caused by all loops . We calculated $\displaystyle L=\frac{n^2\mu_o \pi r}{2}$ considering the winding number fact twice . So shouldn't  be the equation wrong ?  (The question other users are refering as "original" of this question (asked by myself a few weeks ago) isn't totally about self-inductance . This question is regarding to deduction of a equation for self-inductance .)
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A current I flows through a solenoid (with n turns). Self inductance is $L$. If the magnetic flux is $\phi$, I am taught that $-n\phi=LI$  I dont understand the part with $n$, why do we have multiply by $n$? $\phi$ itself represents the whole magnetic flux there, if $\phi$ implied the flux only occurred by one coil, then it would be correct.   Moreover, flux change by a coil ($n$ turns) in $dt$ time is $d\phi$, so the induced electromagnetic force is   $$dE=n \cdot \frac{d\phi}{dt}$$   Why do we multiply by $n$ here? If $\phi$ means the whole flux, then why multiply?
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I am self-studying electromagnetism right now (by reading University Physics 13th edition) and for some reason I always want to understand things in a crystalclear way and in depth. Now look at this simple RL-circuit:    I am assuming the wire is superconducting and suppose at $t=0$ the switch is turned on. My problem is that I want to come up with this differential equation for this circuit: $V-I\cdot R=-L\cdot \frac{dI}{dt}$. I know that I can't use Kirchoff's voltage law $(\oint  \vec{E}\cdot d\vec{l}=0)$, because there is a non-conservative electric field near the inductor. Therefore I use Farayday's law of induction: \begin{equation} \mathcal{E}=\oint_{\Gamma}  \vec{E}\cdot d\vec{l}=-\frac{d}{dt} \int_{S} \vec{B} \cdot \hat{n}\,dA \end{equation}  where $\Gamma$ is a closed loop and $S$ is the open surface attached to the loop. Now, my textbook says:     In general, the total field $\vec{E}$ at a point in space can be the   superposition of an electrostatic field $\vec{E_c}$ caused by a distribution of charges at rest and a magnetically induced, nonelectrostatic field $\vec{E_n}$. That is $\vec{E}=\vec{E_c}+\vec{E_n}$.   Inserting this in Faraday's law I get:  $$\oint_{\Gamma}  \vec{E}\cdot d\vec{l}=\oint_{\Gamma}  \vec{E_c}\cdot d\vec{l}+\oint_{\Gamma}  \vec{E_n}\cdot d\vec{l}=-L\cdot \frac{dI}{dt}$$ Now by definition: $\oint_{\Gamma}  \vec{E_c}\cdot d\vec{l}=0$ since $\vec{E_c}$ is conservative. This gives: $\oint_{\Gamma}  \vec{E_n}\cdot d\vec{l}=-L\cdot \frac{dI}{dt}$, which is perfectly fine and consistent with my textbook. However it also gives (going counterclockwise in circuit): $\oint_{\Gamma}  \vec{E_c}\cdot d\vec{l}=-V+I\cdot R=0$ which it shouldn't. I think my flaw is here, but I've tried to think about this this whole day with no succes.  I have my inspiration from Walter Lewin's video:  but I still can't get the equation by Faraday's law.  Any help is greatly appreciated.
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The Lagrangian density for a single real scalar field theory is \begin{equation}\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)\end{equation} I have often seen this written \begin{equation}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)\end{equation} which appears to differ in the kinetic term.  Specifically \begin{equation}(\partial_{\mu}\phi)^{2}=(\partial_{\mu}\phi)(\partial^{\mu}\phi)\neq\partial_{\mu}\phi\partial^{\mu}\phi=(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\phi\partial_{\mu}\partial^{\mu}\phi\end{equation} where the product rule has been used in the last equality. Is this just sloppy notation (which is acceptable because only first derivatives in the field are considered anyway), or is my maths very bad?
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I don't know how this "paradox" can be solved. I'm given the following system: A permanent magnet with a magnetic field given by ($\hat{a}$ are unit vectors in the x and y directions)  $$\vec{H}=H_0\hat{a}_y$$  and a parallel plane capacitor with an electric field  $$\vec{E}=E_0\hat{a}_x$$  Poynting's vector is given by:  $$\vec{S}=\vec{E}\times\vec{H}=H_0E_0\hat{a}_z \neq 0$$  The funny part comes when the professor told that in a system like that "clearly" there is not propagation (wich I know will imply some short of energy flux) in the $z$ direction, hence the "paradox". Is there or is there not propagation of energy?  Any hint will be appreciated, thank you for your time.
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If we write out Maxwell's equations with magnetic charges, we get $$ \begin{align} \nabla \cdot \mathbf{E} &amp;= 4 \pi \rho_e \tag{1}\\ \nabla \cdot \mathbf{B} &amp;= 4 \pi \rho_m \tag{2}\\ -\nabla \times \mathbf{E} &amp;= \frac{\partial \mathbf{B}}{\partial t} + 4 \pi \mathbf{J}_m \tag{3}\label{Eq:Faraday}\\ \nabla \times \mathbf{B} &amp;= \frac{\partial \mathbf{E}}{\partial t} + 4 \pi \mathbf{J}_e \tag{4}\label{Eq:Ampere} \end{align} $$ In particular, Faraday's law \eqref{Eq:Faraday} contains a minus sign that Ampere's law \eqref{Eq:Ampere} does not. This always struck me as odd because it's often said the fields are dual to each other (i.e. you can replace E with B and &quot;get the same result&quot;), but that requires a bit of mental recalibration to accommodate that minus. So I'm curious what the origin of that negative is and what it means. Are there any intuitive explanations for how to think about it?
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Question Consider two dipoles $({p_1}\hat{i}$ and ${-p_2}\hat{j})$ kept in the $x-y$ plane at $(0,0)$ and $(d,0)$ respectively. Calculate the torque about the COM. Approach 1 Suppose we select the COM as the origin. Then, we label the 4 particles $1,2,3,4$. Let their position vectors be  $r_{1},r_{2}..etc$. Then, when we consider the Force on the $i$th particle due to the $j$th particle= $F_{ij}$. The Torque about the COM.= $r_{i}\times F_{ij}$. When we sum these torques over all particles, and use the following facts:  $F_{ji}=-F_{ij}$ $r_{i}-r{j}=r_{ij}$ $r_{ij}\times F_{ij}=0$  $$r_{1}\times({F_{21}+F_{31}+F_{41})}+r_{2}\times({F_{12}+F_{32}+F_{42})}$$ $$+r_{3}\times({F_{13}+F_{23}+F_{43})}+r_{4}\times({F_{14}+F_{24}+F_{34})}$$ $$= (r_{1}-r_{2})\times F_{12}+(r_{1}-r_{3})\times F_{13} ....$$ $$=0 + 0 ...0$$ We conclude that the net Torque about the COM is is zero. This is somewhat expected, there are no &quot;external&quot; elements in our system. (In fact, this argument suggests that the Torque about any point in the x-y plane is zero. We will still end up with the same expression, as the difference of two position vectors is independent of the choice of the origin.) Approach 2 However, qualitatively, when we use our knowledge of electric fields, we can see that both dipoles seem to experience an anticlockwise torque. They will add up, and thus The net torque is non zero. This appears to be a contradiction. The latter finding is bizarre in the sense that it seems to suggest that the angular momentum of an isolated system is not constant. And the only way that argument 1 can be wrong , is if: $$F_{ij}=-F_{ji}$$ does not hold. I have seen this happen, but the explanation was that the &quot;3rd law of motion&quot; is actually a statement of the conservation of momentum, and in that particular case, some portion of the momentum (of the particles) was carried away by the associated electromagnetic field. I don't think that something of this sort is responsible for my particular case, since there is no $B$ field, and therefore no $S$, which appears in the momentum density of the fields. Apart from this, the two arguments seem to be quite valid individually. So what's going on here? My initial thought was that the fields might be storing angular momentum, but like I said, there is no $B$. Edit 1: I think I have a hypothesis: In approach 2, as the dipoles rotate for some time say $dt$ , they start producing a $B$ field (since we now have moving charges). We now have both $E$ and $B$ fields, so we actually have an angular momentum associated with the field, which perhaps cancels out with the angular momentum produced by the dipoles(=$(T1 +T2)dt$).
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How is the magnetic field at the centre of a loop of current or a solenoid effected by different cross sectional area?  I understand B=unI gives the magnetic field, but it does not specify how it changes as the area of the coil changes.  Thanks.
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I was wondering why Faraday's law of induction and Maxwell-Ampere's law (without sources) are not totally symmetric in the sense that Maxwell-Ampere's law has a $\epsilon_0 \mu_0$ term on the right (in SI units) while Faraday's law doesn't, as symmetry is an important feature in most physical laws.   \begin{align} \nabla\times\mathbf E&amp;=-\frac{\partial\mathbf B}{\partial t} \\ \nabla\times\mathbf B&amp;=\color{blue}{\mu_0\varepsilon_0}\frac{\partial\mathbf E}{\partial t} \end{align}  A popular reference book states the reason being "that we use SI units". Can anyone tell me how using a particular unit can affect the symmetry of physical laws written in their mathematical form?
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The no-load current of a transformer consists of two components:  The Magnetization Current iM is the current required to produce the flux in the transformer core. The Core-loss Current ih+e is the current required to make up for hysteresis and eddy current losses.   This is what my book says for the no-load current in a real transformer. To me it sounded silly that the magnetization current, which is a part of the overall no-load current, is solely responsible for the entire flux generated in the core. How is it that not the overall but a part of the no-load current is responsible for all the flux that is in the core?
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Suppose that there is a semi-infinite wire which extends to infinity only in one direction. There are no other circuit elements at the other end(finite end) of the wire and the current does not loop. The magnetic field obviously has cylindrical symmetry when the Amperian contour is taken as a circle with its center on the wire.   However, due to charge accumulation there is a time-dependent electric field; hence a displacement current. How can I formulate the Ampere's law and show that the magnetic field is the half of that the infinite wire at the finite end of the wire? Do you think treating charge accumulation as a point charge with changing amount of charge right at the finite end of the wire will suffice?   The original question is:
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When t&lt;0 Components in circuit are Jt = 4 A, R1 = R2 = 2 Ω, R3 = 4 Ω ja L = 10 H      At t=0 switch K will be closed and i need to figure out what iL(t) is when t=4  So i tried to figure out iL(t) with  differential equation. First i combined R1 and R2 to get R12=1Ω  Then i transformed power supply from current to voltage Et=Jt*R12=J*1Ω =J Then i combined R12 and R3 to get Rz=5Ω Now i can create equation which is  $$E_t=L*\frac{di(t)}{dt}+R_ti(t)          $$ After adding constant i get $$J_t=10*\frac{di(t)}{dt}+5i(t)          $$ First i found the roots $$10r^t+5r=0 -&gt; r=-\frac{1}{2}$$ So for homogeneous part i get $$y^h=C_1*(-\frac{1}{2})^t$$ And for nonhomogeneous part $$J_t=A$$ $$ \frac{dJ_t}{dt}=0 , \frac{d^2J_t}{dt^2}=0$$ So i get $$  10*0+5=A-&gt;A=5  $$  For complete equation i get $$ y(t)=C_1*(-\frac{1}{2})^t+5  $$ To figure out what is C1 i use t=0 $$ y(0)=C_1*(-\frac{1}{2})^0+5=4 -&gt;C_1=-1  $$ And now for t=4 i get  $$ y(4)=-1*(-\frac{1}{2})^4+5=4.5   $$ I know this answer isn't correct but i don't know what went wrong Also sorry if its little bit hard to read this, i'm new to this
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According to Wikipedia, the  is defined as:  $$Q=2\pi\frac{\mathrm{energy \, \, stored}}{\mathrm{energy \, \,dissipated \, \, per \, \, cycle}}.$$  Here are my questions:   Does the energy dissipated per cycle assume that the amplitude is constant from one cycle to the next. Is it always calculated at the resonance frequency? If the answer to 2 is yes can you explain why for a forced oscillator system with a damping coefficient of $\gamma$ and natural frequency $\omega_o$ the quality factor is $Q=\omega_o/\gamma$ and not some more complicated expression involving the actual resonant frequency (which is not quite $\omega_o$  and is given by $$\omega_r=(\omega_0^2-\frac{\gamma^2}{2})^{\frac{1}{2}}$$ of the system? Is this just an approximation i.e. are we assuming that it resonates at $\omega_o$ but in fact the actual expression is a big more complicated?   Edit: With using the actual value of $\omega_r$ I get: $Q=(\omega_0^2-\frac{\gamma^2}{2})^{\frac{1}{2}}/\gamma$  is this more correct?
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The image of the circuit is shown below and it required to find \$V_0\$,  My first attempt at solving this problem is by changing the current source into a voltage source with 1-V and 2k\$\Omega\$ resistance. The fact that the inverting and non-inverting terminals aren't grounded make this problem look difficult.To the point which, I don't know how to proceed with this question or where to start. I would appreciate any help. Following some thought and another schematic from a hint suggested by Alfred, I produced a schematic representing our work.  And  my solution for the problem is below, Using node equation at nodes A and B we have, $$\frac{V_A-V_B}{1k}=-0.5 \text{mA}$$ $$\frac{V_B-(2+V_A)}{1k}=-0.5-x$$ where x is the current that is sent in the output of the op-amp. Using KCL, at the bottom node near the current source we see that the same current that goes through the op-amp also goes through the \$2\text{k}\Omega\$ resistor. Hence, we have, $$\frac{-V_A}{2k}=x$$ Replacing this in the second equation, $$\frac{V_B-(2+V_A)}{1k}=-0.5+\frac{V_A}{2k}$$ And and solving the equations yields \$V_A=-2 \text{-V}\$ and \$V_B=-1.5 \text{-V}\$
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I was trying to derive wave equation. We know that the curl of an electric field is the rate of change of magnetic flux then why we have to calculate the curl of the curl of the electric field after that. What is the reasoning behind this Faraday's law $$ \nabla\ \times E = \frac{\partial B}{\partial t} $$ Taking curl of the equation $$ \nabla \times \nabla\ \times E = \frac{\partial (\nabla \times B)}{\partial t} $$ I would like to know what is the physical significance of taking the curl of the curl of the electric field.
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In Ampere's law, the current outside the curve taken is not included in the expression. Does it mean that the magnetic field calculated by using the law gives the contribution of only the currents crossing the area bounded by the curve?
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Amperes Law: $\oint\vec{H}.d\vec{l} = I_{enc}$  Is the current, $I_{enc}$, equal to 0 outside of the conductor ($\rho &gt; a$)  because the current density, $\vec{J}$, outside is 0?
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this is my first question on PhysicsSE (I'm already an user of MathSE).   I'm a mathematics students trying to understand Faraday's law, that is   $$\varepsilon= -\frac{d \Phi_B}{dt}$$  where $\varepsilon$ means electromotive force and   $$\Phi_B=\iint \mathbf{B}\cdot d\mathbf{S}$$   means flux of magnetic field. As my textbook points out, there is an interpretation problem here: if the change in magnetic flux is due to movement of the conductor, then free charges in it are subject to Lorentz force, which then causes a current. On the contrary, if the conductor holds steady in a changing magnetic field, the induced current must be explained in terms of an electric field $\mathbf{E}$, described by the equations   $$\begin{cases} \nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0} \\ \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}\end{cases}.$$  Question Do those equations hold if we have a moving conductor in a stationary magnetic field? I guess not: this would mean $\mathbf{E}=\mathbf{0}$. How to solve this?