I want to ask that, is there any intuitive way one can understand that why $\oint \mathbf B\cdot \mathrm d\mathbf l$ across a closed loop is $0$ when a wire not enclosed by the loop has current flowing through it.  In other words why do external currents don't matter?
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I have a question about why this is true  $\oint B\cdot d\ell = \mu_0$ $I_{enclosed}$  My problem is that I am not really seeing visually or mathematically why the magnetic field generated by a current that doesn’t pass through the loop will not contribute to the integral.  Any advice or intuition is aprecciated thanks.
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Say I have an open circuit with a battery. I have a wire connected to the negative terminal and a wire connected to the positive terminal. The wires aren't connected.  In this scenario, would their be electron flow for a very short amount of time? Would the electrons from the negative terminal flow through the wire connected to it, since the potential would decrease between the electrons?  To clarify what I mean, consider the first diagram in this image:    In the first diagram there a large amount of electrons in the negative terminal. Would the electrons in the negative terminal distribute themselves through the wire to minimize the force between them?
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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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Here's an interesting "proof" that there is no such thing as magnetism. I know the answer but I love this so much I had to ask it here. It's a great way to confuse people!  As we all know, $$\nabla \cdot\vec{B} =0$$ Using the divergence theorem, we find $$ \iint_S \vec{B} \cdot \hat{n} \, dS = \iiint_V \nabla \cdot \vec{B} \, dV = 0$$ Since $\vec{B}$ has zero divergence, there exist a vector function $\vec{A}$ such that $$\vec{B} = \nabla \times \vec{A}$$ Combining the last two equations, we get $$\iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ Applying Stokes' theorem, we find $$\oint_C \vec{A} \cdot \hat{t} \, ds = \iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ Therefore, $\vec{A}$ is path independent and we can write $\vec{A} = \nabla \psi$ for some scalar function $\psi$. Since the curl of the gradient of a function is zero, we arrive at: $$\vec{B} = \nabla \times \nabla \psi = 0,$$ which means that all magnetic fields are zero, but that can't be!  Can you see where we went wrong?
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I am not sure which form of  is fundamental, integral form or differential form. Imagine an ideal infinitely long solenoid. When a current is changing in time, can we detect classical effects outside a solenoid, for example generating a circular current around solenoid by Faraday's law. If the differential form is fundamental, we won't get any current, but the integral form is fundamental we will get a current.
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I would like to get some elementary intuition into the problem a solenoid fed with a time-dependent current, and the resulting current that such the solenoid field would induce in a loop completely outside the solenoid.  The classic situation is to imagine a perfect solenoid of $n_1$ turns per meter, having radius $a$ and fed by a current $I(t)=I_0e^{-t/\tau}$.  One places $N_2$ coils (all having radius $R&gt;a$) around the solenoid as illustrated by this picture from Haliday&amp;Resnick 10th Edition.      Using the usual Ampère's law argument shows the $\vec B$ field of the solenoid would be completely contained inside the solenoid, and homogenous across the cross section of the solenoid.    If there is a slowly varying current in the solenoid, will there be an induced current?  Presumably the answer is yes (at least according the HR solution manual): if we take the $\vec B$ field of the solenoid to be changing with time as a result of the current changing in time, but $\vec B$ still homogenous across its cross-section and $0$ outside, there is a change in the flux of this field through the surface bounded by the coils, and even though there is no magnetic field outside the solenoid where the coils are located.  The resulting change in flux induces an EMF independent of $R$, the radius of the coils, or for that matter the shape of the coil, provided it completely contains the solenoid.   If you are suspicious about the use of Ampère's law for slowly varying currents, the same solenoid field is obtained explicitly by Das Gupta in  "Magnetic field due to a solenoid." American Journal of Physics 52 (1984): 258-258, starting from Biot-Savart, which certainly holds for quasi-static currents.     But how can one intuitively grasp that no field outside the solenoid induces a current in a stack of coils located   completely outside this solenoid, in the region where $\vec B=0$?   A "classic" explanation might that the field might be $0$ but the vector potential is $\ne 0$, yet it seems this is invoking a lot of heavy machinery for a 1st year physics problem.  Moreover, Aharonov-Bohm-like explanations are really quantum in their nature and show that in quantum mechanics the potentials are the essential quantities.  Nota: a possible path of solution would be to invoke hidden momentum, along the argument .  Is there a simpler explanation?
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This question is about voltage induced in a inductance.  I know the rate of change of current induces EMF in the inductor. The direction of the induced EMF will be opposite to the change that is causing it. The EMF induced is given by the following formula.  $$V_l = -L \frac{dI}{dt}$$  The negative sign indicates direction of induced emf across the coil is opposite the current change.     In the picture I have drawn current direction and \$V_l\$ direction.  If \$\dfrac{dI}{dt}\$ is postive, \$V_l\$ will be negative according to formula right? Or does the direction of \$V_l\$ has to be reversed?  In many Internet references I found \$V_l = L \dfrac{dI}{dt}\$. Which one is correct?  I hope they reversed \$V_l\$ reference points and removed the negative sign in the formula. Please clarify.
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If a charge travels in a circle it must accelerate, thereby producing EM.  However, a wire in a circular loop is analogous to many charges moving in a circle. So, why don't circular currents produce EM? (I have not found any evidence that circular currents can produce EM).  A similar question is: . However, my question asks about circular wires and direct current. Thanks.
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When the loop is the figure shown below will be rotated in clockwise direction the the slip ring connected to the red segment of the loop will give us the positive voltage and the slip ring connected to the black segment of the loop will give us negative voltage according to the Flemings right hand rule.  What if the loop is rotated in counter clockwise direction ? Will the slip rings provide us the same polarities which they were giving when the loop was moving clockwise ?  Plus  The voltage from the loop is given by the equation.  V(t) = Vmaximum * sin (theta)  Where theta is the positive angle the red segment makes in the clockwise direction or in counter clockwise direction from this position ?
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let us assume that we have a conductor with a specified resistance ( case 1 ) and a normal conductor ( case 2 ) as shown in the figure , and now we apply an external electric field with a battery on it, we know that current density will be different in both the parts ,however , both the parts of the wire have different  resistances as there is a change in its cross section area but the potential difference will be equal as the length of both the parts is equal so is it correct to say that current running through both the parts is equal ?answer for both the cases
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The circuit shown below consists of two resistors connected in series that form a closed loop. No electrical sources are in that loop.Let’s assume that the resistance of the wires that connect the resistors and form the loop is zero.Inside the loop is an external changing magnetic field that induces inside that loop a constant EMF of 1 V.   The current inside the loop is calculated using Ohm’s Law :  The problem was introduced by former MIT professor Walter Lewin. The problem was introduced in this video (starting at 35th min of the video):  In the video is shown that two voltmeters connected to the same terminals show different values. Can two voltmeters connected to the same terminals show different values? My solution to this problem would be: To measure the voltage between points D and A, the circuit has to be divided into two halves, the left half of the loop and the right half of the loop. If the induced EMF is considered to be equally distributed through the loop then in every half of the loop there is an induced EMF of 0.5 V:  The polarity of the the induced EMFs is according to Lenz law.  The voltage between points D and A by writing the equation trough the left half of the loop is:  The voltage between points D and A by writing the equation trough the right half of the loop is:  The conclusion is that there is a unique voltage value between points D and A, and the measured value should be 0.4 V. Now let’s connect to the circuit between points D and A two voltmeters. The two connected voltmeters together with the wires that connect them to the main circuit form a second loop. Let’s assume that the resistance of the wire that connects the voltmeters and forms the second loop is zero. The left half of the second loop is connected to the voltmeter between points C and B, the right half of the second loop is connected to the voltmeter between points F and E. There is also an induced EMF of 1V inside the second loop. If the induced EMF is considered to be equally distributed through the second loop then the final circuit can be represented with the model shown on below.  The voltage between points C and B is defined with the following equations:  The voltage between points F and E is defined with the following equations:  The calculations above match the measurement performed by professor Walter Lewin. My main question is : What is the theoretical potential difference between points D and A?
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I am a licensed radio amateur, and find bewildering the many different explanations, which range from folksy urban myth to Maxwell-Heaviside Equations, of what happens at the termination of a transmission line or feeder. I realise that they all come to the same thing in the end (or should do, pun perfect), but none of them give me a gut feeling for what is going on.  I like diagrams, so an answer in terms of (graphical) phasors for the currents and voltages at the load would suit me best. How, for instance, does a step pulse down the line cause twice the voltage at an open circuit termination? Similarly for current at a short circuit. And how is the reflected step generated by the inductance and capacitance of the line?  Can anyone help, without getting all mathematical, and not telling any "lies to children"?
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Applying the divergence to Eq. $47$, we obtain $$ \mathbf{\nabla} \cdot \mathbf{B} = \frac{\mu_{0}}{4\pi} \int \nabla \cdot \left( \mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right) d\tau^{'}. \tag{48}$$ Invoking the product rule number $6$, $$\nabla \cdot \left(\mathbf{J} \times \ \frac{\hat{\mathbf{r}}}{r^2}\right)=\frac{\hat{\mathbf{r}}}{r^2}\cdot(\nabla \times \mathbf{J})-\mathbf{J}\cdot\left(\mathbf{\nabla} \times \frac{\hat{\mathbf{r}}}{r^2}\right)\; .\tag{49} $$ But $\nabla \times \mathbf{J}=0$, because $\mathbf{J}$ doesn't depend on the unprimed variables, while $\mathbf{\nabla} \times \frac{\hat {\mathbf{r}}}{r^2}=0$, so $$\mathbf{\nabla} \cdot \mathbf{B}=0\;.\tag{50}$$  I have the Fourth Edition of Introduction to Electrodynamics by David J. Griffiths, published by Pearson. I have found this erratum in Chapter 5 Magnetostatics, page 234 . It's just a normal triple product manipulation, i.e. $$ \vec a \cdot \left(\vec b \times \vec c \right) = \vec b \cdot \left( \vec a \times \vec c \right) = \vec c \left(\vec b \times \vec a\right)$$ Therefore, I think that an &quot;equal to&quot; sign is missing between the expressions on the RHS of Eq. $(49)$. I searched for  but couldn't find it.
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Can anyone think of an intuitive explanation of the product rule? I'm not asking for a derivation. To me it seems like quite an un-untuitive result, as apposed to the chain-rule (which is ironically harder to derive).
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Can somebody give rigorous proof that the circle cannot be embedded into the real line? Obviously, if so it must be necessarily continuous and I guess this suffices in this case since there is no continuous injection by intuition...
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Why is the voltage across a capacitor (as a function of time) continuous? Why is the current across an inductor (as a function of time) continuous?
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If we connect an AC source with a pure inductor when AC source is going through 0 volts we get the following graph of the voltage and current.  There are few things in the graph about which i am confused.  In the first graph/diagram "A" of the picture which shows the voltage when it starts to flow and the corresponding current, the current doesnt seem to be lagging behind the voltage by 90 degrees. What is the reason behind that? Is the statement "current lags behind the voltage in an inductor BY 90 degree" is valid just for the steady state voltage shown in graph "B"?  Second confusion is that for the first half cycle of the voltage in graph "A", current is rising both for increasing and decreasing voltage. How is this possible that current is rising in the inductor for both increasing and decreasing voltage?