19
In plain English, it is just Lenz’s law :
Lenz's law, named after the physicist Emil Lenz who formulated it in 1834, states that the direction of the electric current which is induced in a conductor by a changing magnetic field is such that the magnetic field created by the induced current opposes changes in the initial magnetic field.
It is the basic ...
12
The minus sign is what makes Maxwell's equations obey causality, so it's a good thing it's there! To see this, you can write out the source-free Maxwell's equations with the sign of $\nabla \times \mathbf{E}$ reversed in Ampère's Law. If you then to follow the standard construction to extract the wave equation from Maxwell's equations, you would obtain ...
10
Duality is actually not $\mathbf{E}\leftrightarrow \mathbf{B}$ (I've used $c=1$), i.e. $(\mathbf{E},\,\mathbf{B})\to(\mathbf{B},\,\mathbf{E})$. It's $(\mathbf{E},\,\mathbf{B})\to(-\mathbf{B},\,\mathbf{E})$. Defining $\mathbf{F}:=\mathbf{E}+i\mathbf{B}$ is a popular way to check this; the above duality is $\mathbf{F}\to i\mathbf{F}$. It's instructive to ...
7
Piezo electric cells convert mechanical energy to electric energy
7
Doesn't a battery do this? Also, capacitors.
EDIT: With the edit, it looks like the premise of your question could be satisfied by a Van de Graff generator:
https://en.wikipedia.org/wiki/Van_de_Graaff_generator
which uses friction to strip electrons from a substance, and create an electrostatic potential.
5
So we are only putting $mv^2/2$ energy which is gained by the wood but the extra output is current induced in the coil and magnet's motion, why is it not a violation to law of Conservation of energy?
Actually, due to the magnetic radiation reaction force, the energy required in order to accelerate the magnet to $v$ is greater than $mv^2/2$. See https://en....
3
It really comes from relativity, where one uses the field strength tensor:
$$ F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}=\left(\begin{array}{cccc}
0 & E_x & E_y & E_z\\
-E_x & 0 & -B_z & B_x\\
-E_y & B_z & 0 & -B_y\\
-E_z & -B_y & B_x & 0\\
\end{array} \right)$$
When the indices are raised:
$$F^{\mu\...
3
I doubt it because usually all that class of measuring devices basically integrate a periodic (ideally sinusoidal) signal.
In other words, they are conceived to yield an average over time of the amplitude of a signal whose average value varies slowly in time. They might have less inertia than an antique voltmeter or ampere-meter with a physical dial, but ...
3
As a rule of thumb, if you ever transform away a magnetic field the motion of any particle that was due to $\vec{B}$ in the original frame will be due to $\vec{E}$ in the new frame.
This is governed by the lorentz transformation of EM fields.
If a neutral wire had a current going to the right (i.e. electrons traveling left) and you then travel to the right ...
3
To me, Maxwell's equations are synonymous with the work of god, so much so that I'll even take these equations as definitions for the fields. So, what I call the electric and magnetic fields are respectively the two vector fields $\mathbf{E}$ and $\mathbf{B}$ which (in the presence of charges and currents $\rho,\mathbf{J}$) satisfy
$\nabla \cdot \mathbf{E}=\...
2
Faraday's law refers to circulation of electric field along any closed curve in space, this curve does not have to go through conductor, it can go partly or wholly through nonconducting space like air or vacuum or plastic.
So we can apply Faraday's law to coiled wire, just complete the path that goes entirely inside the wire by a path segment that goes in ...
2
Might as well answer the question in case someone in the future is stuck here as well.
The change in flux is 0 but the EMF in the case of an infinite magnetic field should be 0.
That's because while the force you get using Lorentz force law is indeed still there, now you have to take into account the force that the right-most wire is producing, which is in ...
2
Ferromagnetism is explained on a microscopic scale; Within the body of the magnet, the direction of the field lines is explained by the orientation of magnetic domains, and there is no theoretical reason why those could not be oriented to form a "twisted" path. But, outside of the body of the magnet, the direction of the field lines will be ...
2
In vacuum (all $\mathbf{J}$ and all $\rho$ zero) the Maxwell equations imply the wave equations
$$
\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2}=\Delta\mathbf{E}\,,\quad\quad\frac{1}{c^2}\frac{\partial^2\mathbf{B}}{\partial t^2}=\Delta\mathbf{B}\,.
$$
See Vacuum equations, electromagnetic waves and speed of light on Wikipedia.
If you remove the ...
2
Think about it this way: A magenetic field stores energy. Therefore to create a magnetic field you must deliver energy to the system somehow. In the case of an inductor, the only input to the system that can deliver this energy is the electrical power provided to the inductor ($I\times V$).
And if the power provided is not infinite, then the stored energy ...
1
There's an approach from geometric algebra that considers the electromagnetic field as a field of bivectors, and with this geometric interpretation the electric and magnetic components can be rotated into each other. It's too long a discussion to attempt to explain this properly here; read the linked paper if you are interested in the details. I'll just give ...
1
The question is a bit vague, however here is a possible explanation:
There are at least three sorts of light bulbs:
The oldest models with an incandescent filament.
The early economical bulbs which are discharge lamps in disguise (with a discharge tube coiled inside a bulb).
The current power saving bulbs using semiconductor diodes to produce light (i.e. ...
1
If the field is uniform and the velocity is constant then we can pull $\mathbf v\times\mathbf B$ out of the integral.
$$\mathcal E=\oint_C(\mathbf v\times\mathbf B)\,\cdot\text d\mathbf l=(\mathbf v\times\mathbf B)\cdot\oint_C\text d\mathbf l$$
Since $\oint_C\text d\mathbf l=0$, $\mathcal E=0$, so there is no contradiction.
1
The Maxwell-Faraday equation in integral form states that
$\displaystyle \oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} = - \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S}$
On the left hand side of the equation the line integral is around a complete loop and this is called the emf. For the right ...
1
Will the magnetic force become zero in this frame?
It will, for the particle's velocity is zero in this frame.
Will the magnetic field become zero in this frame?
It will not. As long $v \ll c$, where $c$ is the speed of light, the magnetic field in this frame will be practically the same as in the original frame.
if I say that there is no magnetic force, ...
1
The answer here is that you have to consider electromagnetic force as a whole. In special relativity you indeed learn that those 2 forces are actually related to the same phenomenon and when changing from a frame to another, electric field $\vec{E}$ and magnetic field $\vec{B}$ "mix" each other.
In your specific case you'll see the particle being ...
1
An AC current sensor is basically just a transformer. If you just pass the load wire through the magnetic core, it is like having half a turn. Or you can coil the load wire around the core and make it resemble a traditional transformer more for higher sensitivity.
https://www.electronics-tutorials.ws/transformer/current-transformer.html
The load wire ...
1
If you would recall, Lenz's law states that the induced current produced in a circuit always flows in such a direction that it opposes the change or cause that produced it. It's similar to inertia. Now, consider a magnet moving towards a conducting loop along the axis of the loop. Since the magnet is moving closer, the magnetic field experienced by the loop ...
1
They're essentially different scenarios having the same explanation to them. According to Lenz's law, the induced current produced in a circuit always flows in such a direction that it opposes the change or cause that produced it. The change referred to here is a change in the magnetic flux.
Now, when we consider the motional EMF case, the magnet moving ...
1
There are two sorts of e-m induction, to both of which you can apply the equation
$$\mathscr E =- \frac{d\Phi}{dt}.$$
(1) In the first sort, the motion of a conductor gives the charge carriers in the conductor a component of velocity at right angles to the conductor, and in a suitably directed magnetic field there will be a magnetic Lorentz force parallel to ...
1
The picture seems an illustration of the equivalence of the magnetic field generated by the current in the wire ($i_C)$ and the changing of the electric field inside the capacitor ($\frac{\partial E}{\partial t}$).
From the Ampere's Law:
\begin{align}
\oint_{\partial \Sigma} & \mathbf{B} \cdot \mathrm{d}\mathbf{\ell} = \mu_0 \left(\iint_{\Sigma} \mathbf{...
1
the voltage produced is equal to the delta of magnetic field flux over time
Faraday's law says induced EMF (not voltage) is proportional to rate of change of magnetic flux. Voltage depends on other things, such as whether the circuit is open or closed, and in the latter case on effective resistance of the inductor.
If the circuit is open, voltage has the ...
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