I asked why
(a) ##\vec{f}_s=-\vec{E}## inside the battery.
You say this is because
(b) ##\int_-^+ \vec{f}_s\cdot d\vec{s}=-\int_-^+\vec{E}\cdot d\vec{s}##.
But I don't see (b) as a justification.
Of course if I assume (b) is true then (a) is true but this seems to be the result of...
If a battery is in a closed circuit, we have an electrostatic field ##\vec{E}## from the positive terminal to the negative terminal of the battery through the circuit outside the battery. This field generates current.
Inside the battery, the electrostatic field lines go from positive to...
At this point I think I have solved many of my doubts. In this post I will summarize everything.
Consider the following RL circuit
The first thing we do is define a direction for the normal vector to the enclosed surface. This will define the right-hand rule (rhr) circulation direction for...
With this new ##\mathcal{E}_{L,opp}## we have
$$\mathcal{E}_{L,opp}=\oint_{opp}\vec{E}\cdot d\vec{l}=-\oint\vec{E}\cdot d\vec{l}=-(\mathcal{E}+IR)=-\mathcal{E}-IR=L\dot{I}$$
$$\dot{I}+\frac{R}{L}I=-\frac{\mathcal{E}}{L}$$
which gives us the correct solution: a negative ##I## (which means...
I understand the concepts and general calculations but I was hoping precisely to see the calculations for the opposite choice of normal vector and opposite choice of circulation direction. The main issue I have is why do we switch the sign of the term ##-L\dot{I}##?
Suppose we choose the normal...
Here is another attempt to express what I am trying to ask.
Let me write the inductor as a one-loop inductor.
We define the normal unit vector as pointing into the screen.
From the direction of the current through the inductor, we will get a magnetic field pointing into the screen.
The...
Just from looking at this circuit, we would expect the positive current to flow in a clockwise direction.
Let's define the normal vector pointing out of the screen. Then the positive circulation direction is counterclockwise.
I think that means that we have
I am not sure about the...
As far as I can tell we have
$$\mathcal{E}_L=\oint \vec{E}\cdot d\vec{l}=IR-\mathcal{E}=-L\dot{I}=-\dot{\Phi}\tag{1}$$
This differential equation can be written
$$\dot{I}+\frac{R}{L}I=\frac{\mathcal{E}}{L}\tag{2}$$
which is easily solved
$$I(t)=\frac{\mathcal{E}}{R}\left (...
Does this mean we can write the following?
$$\mathcal{E}=\oint_C \vec{E}\cdot d\vec{r}+\oint_C \vec{v}\times\vec{B}\cdot d\vec{r}\tag{3}$$
I haven't seen an equation like the above in my books and notes yet.
What I have seen are two cases.
In one case, we have a uniform magnetic field and we...
I tried to solve this as follows
$$f(t)=(u(t)-u(t-2\pi))\sin{t}$$
$$=u(t)\sin{t}-u(t-2\pi)\sin{t}$$
$$\mathcal{L}(f(t))=e^{0\cdot s}\mathcal{sin{t}}-e^{-2pi s}\mathcal{L}(\sin{(t+2\pi)})$$
$$=\frac{1-e^{-2\pi s}}{s^2+1}$$
where I used the fact that ##\sin{(t+2\pi)}=\sin{t}##.
Then I looked...
I'm going to try to guess my way through the questions I posed.
I'm guessing that since we approximated the solenoid as ##N## loops, then ##C## is just one of the loops.
$$\mathcal{E}_L=-N\frac{d\Phi}{dt}=\oint_C\vec{E}\cdot d\vec{s}\tag{9}$$...
Let me move to my next doubts.
From (8), we have ##\mathcal{E}_L<0##. Given our choice of normal vector ##\hat{k}## in the calculation of magnetic flux, it seems that ##\vec{E}## has fieldlines going clockwise in the solenoid wires.
There is an induced current opposite to the initial current...
That is, a solenoid with ##N## turns, length ##l##, radius ##R##, and a current ##I## flowing.
Let's approximate the solenoid as an infinite solenoid (ie, ##l## is very large).
Then, the magnetic field inside the solenoid is
$$\vec{B}=\mu_0 nI\hat{k}=\frac{\mu_0NI}{l}\hat{k}\tag{1}$$
Suppose...
One thing that confused me was the following snippet
This snippet seems to say that the eddy currents induce a magnetic field that causes the drag force.
But if what I wrote in the last part of the OP is correct, then the induced magnetic fields do not play a role in creating the drag force...
I read this explanation on the Wikipedia entry for "eddy current brake".
Let me go through the reasoning there.
We have the metal sheet ##C## moving to the right under the magnet.
For the parts of the sheet moving toward the magnet, the magnetic flux from the magnet is increasing. From...