The problem would be straightforward if the current $I$ were constant through the toroid. However, this is not the case.  The current is time dependent according to $I = I_{o} e^{-\gamma t}$.  So, if I'm right, a classic Ampere's Law approach doesn't work here. Because the $B$ field through the center of the toroid would change over time, it would cause an electric field, which must be taken into account in the magnetic field.  The toroid in question has a cross sectional radius $r$, internal radius $a$, and outer radius $b$.   I feel like I am missing something, or approaching the problem incorrectly, as I do not know how to find the $E$ field induced by the changing $B$ field. From there, I do not actually know what the math would look like. I also have to calculate the self-inductance of the toroid after this, which I would do by finding the flux through the cross section of the toroid and comparing it to the current through the same area. But, I can't do that yet, because I don't have the magnetic field to find the flux!
121906
Say you have a current $I(t)$ (notice the time dependence) flowing through a Toroid with $N$ total loops and all the usual approximations: $(b-a) \ll r,\; B=0$ outside.   (source: )    You are asked to calculate the magnetic field $B$. Could you apply Ampere's Law to obtain the familiar form of $B$ found for example ?:  $$B~=~\frac{\mu N I}{2\Pi r}.$$
100113
Remark: This is not a homework question...It is pure out of theoretical interest. I asked this the mathematics-community a couple days ago and got no answer, so I figured I'd try here.  Most standard physics textbooks compute the force two infinite wires exert on each other, but they remain silent about the case where the wires are finite. Let's say we have two parallel wires carrying a current of equal magnitude in the same direction, both of which have a length $d$ and also seperated by a distance $d$. I now want to find out the force one wire exerts on another, using the Biot-Savart Law.  Let the left wire be positioned at the origin of the $xy$-plane, going along the $y$-axis, and let the other wire be a distance $d$ to the right. We assume the currents are flowing in the positive $y$-direction. Then we first choose a source element (on the left wire) of infinitesimal length $dy$ described by the position vector $\mathbf{r_0} = y_0 \hat{j}$. This constitutes a current source of $I d\vec{l} = (Idy) \hat{j}$.  We then pick an arbitrary field point $P$ on the other wire with position vector $\mathbf{r_p} = x\hat{i} + y \hat{j}$. Then the position vector $\mathbf{r}$ pointing from the source point to the field point is given as  \begin{align*}  \mathbf{r} = \mathbf{r_p} - \mathbf{r_0} = x\hat{i} + (y - y_0) \hat{j}, \end{align*} with $\sqrt{x^2 + (y-y_0)^2}$ being the length of this vector. If we now calculate the crossproduct $d\vec{l} \times \mathbf{r}$, we can write it as \begin{align*} dy \hat{j} \times (x\hat{i} + (y - y_0)\hat{j}) = -dy x \hat{k} \end{align*} Now comes the tricky part. I think I need to setup a double integral, because we are working with infinitesimal force elements $d\mathbf{F}$, each which is given as  $d\mathbf{F} = I d\vec{l} \times \mathbf{B}$. But we also have that \begin{align*} d\mathbf{B} = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \hat{r}}{r^2}  = \frac{\mu_0 I}{4 \pi} \frac{d \vec{l} \times \mathbf{r}}{r^3} = -\frac{\mu_0 I}{4 \pi} \frac{dyx}{\sqrt{x^2 + (y-y_0)^2}} \hat{k} \end{align*} Hence I need to somehow integrate over $d\mathbf{F}$ and $d\mathbf{B}$. Does anyone have an idea how to do this?
61129
I am happy to always not understand how does a transformer works ... One answer give me 10 questions ... (So I suggest you to not answer to this question, if you do not want to have more questions :D)  Here is the circuit :     So I put the "coupling" factor to 1, in order to have no leakage inductance and I set the value of the inductors very high in order to have a very low magnetizing current. So the primary current is approximately equal to the secondary current. My question is the following : How the primary current can reach an equilibrium to be equal to the secondary load current ? I will not accept answer which tells Pout is equal to ... So Pin is equal to Pout... There is something happenning into the transformer ...   Thank you very much and have a nice day !
21655
I'm trying to get my head wrapped around transformer operation and in the process regretting the times I snoozed in my Electromagnetics class as a EE student back when I was a lad :)  I'm looking for an intuitive understanding, but not just an analogous one. I'd like it to be grounded in the actual physics of what's happening. I've found several excellent sources on the web, but they all seem to skirt this question.  I've come across a few interesting hints and am now tantalizingly close, I think, but still yearning :)  Fact 1: Although varying sinusoidally, the "peak-to-peak" flux, so to speak, in a transformer's core is essentially constant (for a given voltage applied to the primary), regardless of the load.  My intuitive hypothesis was that variation in the "strength" of the flux was what transferred the energy from the primary to the secondary, but this fact would seem to contradict that theory. I had thought that the primary makes a bunch of flux based on the current flowing through it and the secondary sucks it up to make current of its own. No dice, it seems.  Then of course there's the fact that the formula for flux involves only voltage, time (frequency), and turns :)  Fact 2: The current in the primary is (approximately) 90 degrees out of phase with the voltage at no load, and approximately aligned in phase at full load.  This fact seems very promising and also curiously satisfying. It would imply that the Volt-Amps (VA) of the primary is constant and only the power factor changes as the current load on the secondary increases.  But I still don't get how the energy is actually being transferred. It seems vaguely like the flux is just there as an energy conductor or something and some other phenomenon is actually doing the energy transfer bit.  Can someone see what I'm missing and explain what's actually happening in there?
41466
I have a conductor with volume $V$, passing a magnetic field($B$) with velocity($v$):   I'm trying to calculate the Eddy currents to figure out the magnitude of the drag force($F_d$) generated the Eddy's, how can that be done?    I could not find a direct formula, I noticed from a lot of sources that it seems to be a difficult task.  The conductor is in an open circuit(I don't think it makes a difference if it was in a closed circuit). Please let me know of any additional information I'm missing(not sure of all the required variables). I'll add them as an edit.
22648
I have been revising basic compound angles and I am struggling to understand the following question from the examples I have previously studied on such topic.     I cannot envisage the compound angle formulae associating with this (if I am correct in saying it associates with this topic).     A first step, or point of direction/suggestion would be brilliant. Thank You.  The instantaneous current $i$ and the instantaneous voltage $v$ in a pure resistance a.c. circuit is given by;  $i=i_{\text{max}} \sin \omega t$     and     $v= v_{\text{max}} \sin \omega t$  Since power;  $P = iv =i^2 R$  Show that an equation for instantaneous power is;  $P=i_{\text{max}}^2 R[1-\cos 2(\omega t)]/2$
75511
In context to the diagram below , it is quite easy to find the current flowing through section AD , which turns out to be 2A , however , I cannot determine the current in any other branches with Kirchoff's Voltage Law . The branches AB,CD,DE and FA have no resistance across them so whenever I take those branches in any of the the possible loops , the potential drop/rise will always be 0 so the terms involving 'I' never appear and consequently they are indeterminable.So how do I find the magnitude of 'I' ? In this particular case I can guess than I=1A with junction rule and the symmetry of the circuit - the batteries have same emf, so what if the emfs aren't same ? then the original problem comes up again.
64980
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
36568
So I started off with electrostatics and everything seemed nice and mathematical and justified and then "DC circuits" happened!  I just cannot understand the model of electron flow in electrical circuits.  Here are my specific doubts-:  1) If potential difference across a tiny cross section of conducting wire is zero, then why on earth does electron flow across that cross section at all? Never mind potential difference across the whole circuit.   2) Is there a constant electric field across a wire connected to a battery? If yes then how is potential difference across a zero resistance wire constant?   Shouldnt it be increasing?  Doesnt it violate ohms law?  If no,  then why do electrons flow at all?  Please take time to consider these doubts and relieve me of my frustration.  I havs searched through the net for this but every answer seems like beating around the bush.  All of the 4 books I have consulted do not address these facts to my satisfaction.   Frankly I think nobody understands this.
91514
We know for a metallic conductor   Current $\frac{I}{enA} = v$ where $v$ is drift velocity , $e$ is the charge of an electron, $n$ is no of electrons per unit volume and $A$ is area of cross section.  Ok my questions that I am struggling with:   Does it depend on length of wire? (let's say if the current is the same and area is also the same but one wire is longer and one wire is shorter then is the drift velocity for both wires the same?) Does it depend on the cross sectional area of the wire? (Yes, according to the relation, if the area is greater, drift velocity is lower, correct?)   I have digged university physics still I couldn't get my answer :( .
74795
We know changing magnetic flux induces electric field which makes current to flow .here in the below picture the flux is changing through only circular coil bout not through the rectangular part.so what induces electric field in rectangular part for current to complete the circuit ..Note that the electric field is required inside awire to make the current flow.
103133
In the usual bookwork treatment, it is easy to show that the differential and integral forms of Maxwell's equations are equivalent using Gauss's and Stokes's theorems. I have always thought that neither version is more fundamental than the other and each has its place in solving problems. (See also  )  But: I have a conceptual problem with applying the integral forms of these equations in cases where there is time-dependence and the "size" of the loop or area means there is a significant light travel time across the regions considered compared with the timescale on which fields vary.  e.g. Suppose there is a time-varying current in a wire $I(t)$ and I wish to find the fields a long way from the wire. My first instinct is that this ought to be solved using the inhomogeneous wave equations to give A- and V-fields that are dependent on the retarded time - hence leading to the E- and B-fields.  But what about using Ampere's law in integral form? What is the limit of its validity? If we write $$ \oint \vec{B}(r,t)\cdot d\vec{l} = \mu_0 I(t) + \mu_0 \int \epsilon_0 \frac{\partial \vec{E}(r,t)}{\partial t} \cdot d\vec{A}$$ then presumably the $t$ that is defined on each side of the equation cannot be the same, since a change in $I$ at time $t$, presumably leads to a change in $B(r)$ at a time $t + r/c$? I suppose one does not care about this so long as the timescale for a current change is $\gg r/c$.  So my question is: Are the integral forms of Maxwell's equations inherently limited by this approximation, or is there a way of formulating them so that they take account of the finite size of a region in cases where the fields are time-variable?
41239
So I have been mulling over Ohm's law for hours today. It makes sense to me that the current gets bigger as voltage rises, since the voltage is supplying more electrons and current is the number of electrons flowing. However, I'm confused by the theory that as resistance drops, current increases, even when voltage stays constant. Such as:  $$V/R = I$$ $$\frac{1V}{0.0001 \Omega} = 10,000,000 mA!!$$  That's a big current from such a small voltage!  My mental picture of a current is more electrons flowing. Electrons are supplied by the voltage, right? If we start off with a small voltage, say 1V, but infinitely decrease the resistance, we'll get a bigger and bigger current. How can this be? Isn't current the number of electrons flowing? With fewer electrons (because of less voltage), how can less resistance infinitely increase the current?
109036
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.
89382
In the  (semi-classical, but should still apply here I think), the conducting electrons are in a constant electric field, and, in between collisions with the lattice ions (that happen, on average, in a time $\tau$ since their last collision) in which their velocities are reset to an average of 0, they are accelerated and gain a momentum  $$p = qE\tau (=mv)$$, where $q$ is its charge and $E$ is the electric field from the applied voltage. This obviously means in between the collisions, the electron accelerates $a$, from $v_f = v_i + a\tau$.  Accelerating charges give off . So does a DC carrying wire give off radiation one can measure?  I suspect it does, but it's of an insanely small magnitude. A quick order of magnitude calculation with the following values gives me:  $$q \sim 10^{-19} \,{\rm C}, \\ E\sim 1\,{\rm V}/1\,{\rm cm}=100\,{\rm V/meter},\\ m\sim10^{-30}\,{\rm kg} \\ \Rightarrow a = qE/m \sim 10^{13}{\rm m/s}^2 \\ \Rightarrow P \sim q^2a^2/\epsilon_0c^3 \sim 10^{-28}\,{\rm W}$$  Which is obviously pretty small. I'm not sure what's the smallest power that can be measured, but I'm guessing by increasing the applied voltage one could increase the radiated power. But is the idea even fundamentally sound?
119375
I am an electronics student, and I was set some homework online. I am not having trouble with the question itself but the theory. It says that the current through the resistor is the same as the maximun current that the diodes can safely handle, but as you can see, there are (\$V_{out}\$) volts being dissipating to the right, and a minimum required current going through the diode to maintain consistent voltage.  The thing is, how could the current through the resistor be the same as the current through the diode, as \$Voltage=J/C\$, or, joules per coulomb, so that means there must be some energy carried by the coulombs, and ultimately, some electrons. But then this is contradictory, wouldn't the current going through the resistor be the total current, the current dissipated to the right and current through the zenor? Or... is it that there isn't, then, any current passing at \$V_{out}\$? and it doesn't represent voltage being dissipated? I cant wrap my ahead around it...
33954
Suppose you place a point charge $+Q$ at the corner of an imaginary cube.  Since electric field lines are radial, there is no flux through the three adjacent (adjacent to the charge) sides of the cube. However there is some amount of flux passing through the other three sides of the cube (flowing out of the cube).  We can estimate that the flux through these three surfaces combined is equal to $Q/(8\epsilon)$. As, if you consider $7$ other cubes having the charge at the corner, each of them would have equal flux flowing out by symmetry and since the total flux through the $8$ cubes is $Q/\epsilon$, each cube would have a flux of $Q/(8\epsilon)$.  Now apply Gauss' law to the cube, and we find that the cube encloses a charge of $Q/8$.  This means that 1/8th of the charge belongs to this cube.   But the charge we placed was a point charge with no dimensions. It cannot be split into parts.  What is wrong?