35
It looks like an induction coil with the make and break device at the bottom and a switch right at the bottom. If you connect it up to an accumulator, be very, very careful as the output between the two balls, when separate, could be lethal. Also the electrical insulation elsewhere may be poor and you might get a shock just by touching the switch.
Use ...
25
The physics you are looking for is electromagnetic induction (https://en.wikipedia.org/wiki/Electromagnetic_induction#Electrical_generator)
When you move a permanent magnet relative to a conductor (the copper wire), the magnetic field of the magnet influences the electrons in the copper, creating a current. Really, the energy that you put in to the system by ...
22
The fan motor provides a torque $\tau$ which has to accelerate $\alpha$ the fan blades whose moment of inertia is $I$:
$$\tau=I\alpha$$
Given how long it takes for the fan blades to stop the frictional torques must be fairly low and so the torque applied by the motor to keep them going must also be low. With the relatively small torque rating, even if the ...
22
The definition of magnetic flux is
$$\Phi = \int_S d\vec{A}\cdot\vec{B},$$
where the integral is not over a closed surface in general. Gauss' Law requires that the integral is over a closed surface, and so there is no contradiction.
In particular, look at any basic discussion of Faraday's Law. They always look at simple loops or coils of wire. There are ...
21
It is a spark radio transmitter.
The first working radios.
Video: https://www.youtube.com/watch?v=YSf93g0heUA
Pics: https://www.google.com/search?q=spark+radio+transmitter&source=lnms&tbm=isch&sa=X&ved=0ahUKEwi-68m5vJjfAhXMx1kKHVuUASQQ_AUIDygC&biw=1920&bih=930
This one looks awfully similar and might give you some help finding ...
20
Koldrakan’s answer explains how the energy is generated. But you might be confused as to why the bulb keeps glowing for some time rather than the light itself fluctuating with the shake.
If you didn’t already know this, that's due to the capacitor. A capacitor can store energy in the form of charge. When you shake it the electric energy generated gets ...
20
Practically speaking yes, this will almost certainly be a step-down transformer, but I would agree with those other teachers: it can't really be concluded from the wire thickness. It's perfectly possible to build a transformer in which the secondary coil has more windings, but nevertheless use thicker wire.
As John Rennie explained, this doesn't normally ...
15
An electromotive force doesn't require a conductor --- it doesn't even require matter. The electromagnetic field is a local property of the vacuum, governed by Maxwell's equations. The relevant one in this case is
$$
\vec\nabla \times \vec E = -\frac\partial{\partial t}\vec B
$$
That is, at any point in space, a changing magnitude or direction for the ...
14
The thickness of the wire determines the maximum current the wire can carry without overheating. Thicker wire means a greater current.
With a transformer the power coming into the primary, $P_p = V_pI_p$, is the same as the power coming out of the secondary, $P_s = V_sI_s$, (less a few resistive losses) and this means $VI$ is constant.
For a step down ...
11
A much simpler way of thinking about this is to consider energy. When the fan is spinning it has quite a lot of kinetic energy (try to stop it by putting your finger in the way to confirm this (don't actually do this!)). That kinetic energy goes as the square of the rotation rate, in fact.
So as the fan starts, the motor needs to add energy to it. It ...
10
You might be thinking in comparison to a desk or handheld electric fan.
As mentioned by @Farcher, $\tau = I\alpha$. $I$, the moment of inertia of a spinning body around a particular axis of rotation, is calculated as follows:
$$I = \iiint\rho(x,y,z)||r||^2\ dV$$
Or with uniform density,
$$I = \rho\iiint||r||^2\ dV$$
From this formula, you can see that ...
9
I do not think that I need to draw a closed surface but here is an example of an open surface with a closed loop at its throat.
This is likened to a butterfly net.
It is often the case that the closed surface is taken to be in the plane of the loop for ease of calculation but this does not always have to be so.
The answer to a recent question illustrates ...
9
The transformer does draw a current. But because the unloaded primary coil is a reasonable inductor, the current is almost 90 degrees out of phase with the voltage, so that the total power is low.
8
It's a bit complicated (Wikipedia). Induction motors work in sync with the AC frequency but have no torque at 0 RPM so they need some arrangement to get them started.
8
When we have a DC voltage source with a switch in series with RL and the switch is closed at t=0 then it is said that current is zero initially, but the voltage across inductor is same as that of applied voltage( according to kirchhoff voltage law) so there should be current( according to v=L(di/dt) )but it contradicts the initial statement so how do I ...
6
Neither Coulomb nor Biot-Savart are correct equations for the electromagnetic field except in statics. There are time dependent generalizations, such as Jefimenko's equations.
$$\vec E(\vec r,t)=\frac{1}{4\pi\epsilon_0}\int\left[\frac{\rho(\vec r',t_r)}{|\vec r -\vec r'|}+\frac{\partial \rho(\vec r',t_r)}{c\partial t}\right]\frac{\vec r -\vec r'}{|\vec r -\...
6
Gauss's law states that $\int_S \vec B\cdot d\vec S=0$ for a closed surface, while the induction law relates the flux through an open surface to the electromotive force ($\xi$) in the circuit formed by its border
6
In a transformer the thicker wire is usually the one that carries a larger current. According to the simple formula relating the voltages and currents in a transformer $V_1 I_1=V_2 I_2$, you have the larger current at the lower voltage terminal. Thus if the primary voltage of this transformer is the mains voltage, then this is a step down transformer to a ...
6
For a stack of coils, as you say, this is actually a famous result, but not quite in the way that you've phrased it.
If you drop a magnet through the air from some height $h$, it'll generally hit the ground with a speed $v$ that obeys $\frac12 mv^2 = mgh$ --- that its, its gravitational potential energy gets converted into kinetic energy.
However if you ...
5
You can actually arrange a simpler experiment for this. Suppose that Lenz's law were reversed and induced currents reinforced the change of the magnetic flux. Now take a single loop of wire, and suppose that we produce a small current in it (with a battery or a magnet - it doesn't matter). An increasing magnetic flux is then created through the loop.
This ...
5
The third of Maxwell's equations (Faraday's law) says that a changing magnetic field has an E-field curling around it. The closed line integral of this electric field is the EMF that drives the induced current in the conducting wire. At a microscopic level, the curling electric field, which has a significant component parallel to the wire, exerts a force on ...
5
The break in the circuit will stop electrons from flowing, but it does not stop the EMF from being induced. The EMF is simply canceled out by the electrostatic forces which prevent the electrons from "bunching up".
Without the flow of current, there is no generation of a secondary magnetic field, which is the thing which would normally slow the fall of ...
5
What matters is the rate of change of the magnetic flux through a closed loop (or an open surface):
$$
{\cal E}=-\frac{d}{dt}\int_C \vec B\cdot d\vec S
$$
It doesn't matter how this change in flux is produced.
You can start, for instance, with a static magnetic field produced by a wire carrying a constant current, and drag a square frame at constant speed ...
5
No.
Any static electric current will generate a magnetic field which is not attributable to a non-conservative electric field. I'm unsure why you haven't been exposed to that example, but you can find it in the introductory section of the magnetism chapter of pretty much every textbook on electromagnetism.
5
This is where it's a good time to "converse with the math". Let's look at the equation:
$$\mathcal{E} = \oint_\gamma \mathbf{E} \cdot d\mathbf{l}$$
which in this case equals $\mathcal{E}_\mathrm{ind}$. This is based on the work formula:
$$W = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$
Thus, what your question is, is essentially, asking "how can you have ...
4
Of course it is the same. There is no preferred system of reference, and if you sit in a train that is passing by the experiment with 100 mph, you would still see the same thing.
Just imagine that you are tiny and sit on the magnet, and the 'moving' magnet is not moving for you (and actually, you sit on the earth, and it is moving).
The frame of reference ...
4
The difference between the solenoid and the magnet is that the magnetic field in the soleniod is a transitory thing, created by our motion of the second magnet. In the first case, the fixed magnet has it's own magnetic field and the system already has energy in it - put there when we first moved the magnets apart (or created the second magnet). We can get ...
4
Notice, the magnetic magnetic field $B$ at the center of a coil carrying current $i$, with radius $r$ & having $n$ no. of turns $$B=\frac{\mu_0}{2}\frac{ni}{r}$$
hence, magnetic flux $\phi$ linked to the coil is given as $$\Phi=BA=\frac{\mu_0}{2}\frac{ni}{r}\pi r^2=\frac{\mu_0 \pi nir}{2}$$ now, setting the value of $\phi$, we get
$$L=\frac{n\Phi}{i}=\...
4
You can't get away from special relativity, which is what "unifies" electric and magnetic phenomena. The electromagnetic field really is a single field (actually a tensor, but don't worry about that), and its components mix together in the Lorentz transform in a similar way to how the x,y,z components of a usual vector field mix together in a rotation.
...
4
TL;DR: The magnet will asymptotically approach a "terminal velocity", at which the magnetic force from the currents in the walls of the tube is exactly balanced by the force of gravity.
To see this, let's denote $F_\text{Lenz}$ as the force from the currents in the wall. This force will obviously depend on the velocity $v$ of the magnet. Its magnitude ...
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