25

First, Field strength. This calculation is strictly an electric potential calculation; radiation and induction are safely ignored at 50Hz.* For a 200kV transmission line 20m above ground, the max electric field at ground level is about 1.2 kV/m.** This number is reduced from the naive 200kV/20m=10 kV/m calculation by two effects: 1) The ~1/r variation ...


20

The energy is not used up. It goes to the magnetic field, and when the magnetic field is at its strongest value there is no energy left in the electric field of the capacitor. But then the magnetic field starts decreasing as the capacitor charges back up because the current starts decreasing. And when the capacitor is fully charged there is no current and no ...


16

Capacitors and inductors are images of one another under the self-inverse mapping that transforms a linear electrical network to its dual network. The network duality transformation maps the network's graph to its topological dual graph,then all the impedances (either as lone-frequency complex scalars or as Laplace transfer functions) in the dual graph ...


12

There is no analogy of "voltage", "current", "charge" or "flux" to electromagnetism for the weak force, at least none that would be helpful. The reason for this is that all of these are classical concepts, while the notion of the weak force is completely quantum. Taking the classical limit just makes it vanish because the classical force law of forces with ...


12

It depends as to whether you are considering the LC(R) circuit in isolation or with a alternating voltage supply as the driver. The series of diagrams illustrate what might happen if the capacitor is initially charged and then connected to an inductor (with resistance). So what is happening is essential an exchange of energy associated with a magnetic ...


9

If the power line is 20m high, and has the voltage of 1MV , then the electric field (near ground), very roughly, is on order of 1000/30 kv ~ 30 000 v/m (the numbers are very approximate and the field is complicated because it is a wire near a plate scenario, and wire diameter is unknown but not too small else the air would break down, i.e. spark over, near ...


9

In the limit of long times, the currents are steady, so the magnetic fields they create are steady so there is no induced EMF. This situation is usually tagged "steady state". That said, there will be a period of time when you have just switched a circuit on or off during which things have not settled down and then there will in general be effects not seen ...


9

In this answer, I'm going to use $Q$ instead of $q$ for the capacitor charge. There is only one certain rule for finding Lagrangians: The Lagrangian is chosen such as to get the correct equations of motion. Never forget that. In the case of a circuit problem, the most sure way to know you got the right Lagrangian is to see if it gives you the right ...


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 ...


7

It's a reasonable question, and the answer is: one can't prove it, without introducing induction. Consider a conducting loop with no current. Then someone starts creating a current in it, using, for example, a battery. The question is: why should we perform work to create this current, if we know that magnetic force $$ \mathbf {F} = \frac{q}{c}[\mathbf{v},\...


7

The original question talked about a discrepancy between the result obtained by calculating the flux directly and using the definition $L= \Phi/I$. The confusion arises because of a concept known as "flux linkage". When you calculate the flux enclosed by the region of unit length between r and r+dr, you calculated an expression for flux which you integrated ...


7

There does not need to be a magnetic field in the inductor for there to be "back emf" (I would prefer "induced emf"). The induced emf is the consequence of a changing magnetic field and not of a magnetic field itself and hence there can be a changing magnetic field even at zero magnetic field (something like a positive acceleration downwards for a ball ...


7

My book says the current through the inductor would pick up to 0.2A, while the current through the capacitor drops to 0A. This is correct. To find the DC steady state solution for this circuit, replace the inductor with a (ideal) wire and replace the capacitor with an open-circuit. Why? In DC steady state (the solution as $t\rightarrow\infty$), all the ...


6

Transformers generate oscillating magnetic fields at the mains frequency and the fields produce an oscillating force on: anything nearby that's ferromagnetic (like the core) anything nearby that is carrying a current (like the windings) The sound you hear is because various bits of the transformers are moving in response to the oscillating fields and this ...


6

If the emf due to the solenoid is assumed to oppose the applied voltage and have equal magnitude (in volts), there is zero electromotive intensity in the wire. Since current is assumed to be present, this means the current flows even while total electromotive force vanishes. This is possible for wire made of perfect conductor (superconductor). In practice, ...


6

however I can't think of the fundamental reason as to why it is completely impossible to charge a capacitor with anything but 50% of the battery energy whereas an inductor could theoretically store 100%. Essentially, to charge a capacitor with finite current from a voltage source requires dissipation. Why? Consider the ideal capacitor equation (in ...


6

Therefore, current will increase, and voltage will also increase across the entire circuit. It's not clear exactly what you mean by "across the entire circuit", but think about your model of a battery. For a simple analysis that is usually used for circuits like this, the battery is considered as a constant voltage source. Therefore, by Kirchoff's Voltage ...


5

An induction cooker works with a flat coil setup similar to the one shown below. The diameter of this coil depends on the exact model (in the range of 30 cm). This implies that the field is also extending quite a bit into the air above the cooker. As the large field can be potentially hazardous (think metal ring on finger) there are relatively strict ...


5

I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn. 1. Why self-inductance is not considered when solving Faraday's law problems Self inductance should be ...


5

There's a really awesome trick for problems like this. This is going to be a long post but the method presented makes problems like this really easy. The idea is to turn the series branch $C_2$, $R_2$ into an effective parallel $R$ and $C$. See the diagram. The effective parallel values are denoted $C_{2,p}$ and $R_{2,p}$. Parallel capacitances just add, so ...


5

Does that mean when I apply a voltage, the current will be infinite large? No, not even in the context of ideal circuit theory. It's a bit subtle since we're using phasor voltages and currents and that requires a couple of assumptions to hold in order to be valid. When those assumptions don't hold, we have to see what the 'infinity' (division by zero) ...


5

Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You can ...


5

If you have a single tube, the current will flow on it directly without making the $N$ loops. It will result a different direction, i.e. different magnetic field, its magnetic field will be much weaker. Having the loops, the magnetic fields created by the induvidual loops is added. Actually, you have "the same current" using $N$ times to produce the field. ...


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

Since you want to avoid differential equations, I will instead consider the so-called Phasor domain, which is actually nothing but the Fourier transform of the original signals. In phasor domain, we will be basically considering complex values: Complex voltages, complex resistances (which are denoted by $Z$ and called impedance): This is simply the ...


5

It's not really true to say that a capacitor stores an electric field, or an inductor stores a magnetic field, though there are certainly fields associated with both. I think a better way of putting it is that both capacitors and inductors store electromagnetic energy. In that case your question becomes: Why do we not have devices like inductors or ...


4

I know this post is old and has been answered I thought I would post the exact derivation to help out anybody in the future. In order to calculate the the internal inductance of a wire we have to equate the equation for the energy of the magnetic field to the energy from the inductor/inductance. Energy of the $B$ field: $\frac{1}{2\mu}\int B^2dV$, where $...


4

why do we not get a shock? Because the electric resistance of a human body is by orders of magnitude higher than the resistance of the steel pot. why is it that current is converted to heat while it has a good conductor(say, steel) to flow through? According to Maxwell–Faraday equation, changing magnetic field creates the electric field, i.e. the ...


4

I believe you simply need 4 op-amps. Here is what you need for differentiation and integration. http://en.wikipedia.org/wiki/Operational_amplifier_applications#Integration_and_differentiation so you simply amplify u by factor 2 with an opamp-circuit then use 2 op-amps for differentiation and integration at the end you should add them all with an opamp ...


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