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23

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 ...


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 ...


9

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 ...


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 ...


8

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 ...


6

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 ...


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 ...


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 ...


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

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 ...


5

Edit 1: I think I just understood you question: you are actually trying to calculate some sort if “internal” inductance, i.e. the contribution to the inductance of only the field inside the conductor. When calculating the flux, you have to choose a closed path over which you would want the electromotive force, and then integrate the magnetic flux over the ...


4

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 ...


4

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 ...


4

Have a look to this old paper: http://www.g3ynh.info/zdocs/refs/NBS/Rosa1908.pdf


3

When calculating the volts per meter of the static field, it's important to assume that the bicycle is conductive (presumably an aluminum frame). Without the bicyclist, one would use image charges to calculate the electric field at the bicycle. The three phases should partially cancel, and Art Brown's calculation seems reasonable, around 1200 volts per ...


3

The relationship $\vec E_c + \vec E_n = 0$ does not hold globally so the two fields are not scalar multiples. Within, and only within, the (ideal) conductor that forms the inductor, the electric field must be zero. Note that the text specifically says so the total electric field ... within the coils must be zero. UPDATE: Consider the general $E$ ...


3

Given $\theta = 30$ and pure resistance is $20 \Omega$. Using a phase diagram it could be observed that $ \tan{\theta} = \frac{\chi_L}{R} $. Find $\chi_L$ and substitute that into $\chi_L = 2 \pi f L $. It's pretty straightforward I believe?


3

Consider an ideal transformer for simplicity (from Wikipedia). (the voltages and currents shown in the picture are phasors) If you consider the above circuit (transformer) as a block in a larger circuit, because it is an ideal transformer you will not have any energy loss in the block. So you can write: $$I_P \times V_P + I_S \times V_S=0$$ and also, ...


3

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 ...


3

"Mutual inductance" and "self-inductance" are not the same thing, no. Suppose you have two loops (of wire, for example), loop A and loop B. When you run a current through loop A, it produces a magnetic field that fills the space around both loops. The mutual inductance between the two loops, denoted $M_{AB}$, is defined as the ratio of the magnetic flux ...


3

When the switch is opened, the inductor energy is dissipated over time via the internal inductor resistance and the external parallel resistance. The stored energy is radiated away as heat, heating both the inductor and the external parallel resistor. If the external resistance is much larger than the internal inductor resistance, the majority of the ...


3

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 ...


3

Here is an approximation of what you are trying to do. The magnetic field inside a solenoid is given by (from Wikipedia) $B={\mu}nI$ where ${\mu}$ is the permeability of the medium (presumably water in your case), $n$ is the number of turns per unit axial length, and $I$ is the current. The maximum value of $n$ is determined by the thickness of the wire you ...


3

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 ...


3

Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy ...


3

For a slightly more prosaic elucidation, again using the electrical circuit paradigm, I find it helpful to rework everything into a (rather approximately, surely) "pseudo-Ohm's Law" or "impedance-oriented" configuration: Inductance: $$V=\frac{\textrm{d}I}{\textrm{d}t}\cdot L$$ Resistance: $$V = I\cdot R$$ Capacitance: \begin{align}\Delta V& = \Delta ...


3

As you slide the switch open, it doesn't instantly transition from zero resistance to infinite resistance. As the contact area decreases the resistance rises, which acts upon the current to produce a voltage operating against the current. Even when the contact resistance becomes zero, there is capacitance across the gap, which produces a rapidly rising ...


3

First of all, don't mix up voltage with current. In your examples 1 and 2 it is certainly true that the voltages across the resistor and inductor are the same w.r.t. the source voltage. This is just Kirchhoff's voltage law. However, this still results in a current lag in the inductor compared to the resistor. Say the source voltage is $\Delta V_S = V_0 ...


3

It seems entirely symmetrical. But it isn't symmetrical. For the voltmeter on the right, the 'outer' loop that encloses the magnetic field consists of the voltmeter, leads, and a $100\Omega$ resistor For the voltmeter on the left, the outer loop that encloses the magnetic field consists of the voltmeter, leads, and a $900\Omega$ resistor. Moreover, ...


3

The energy stored in a field is the energy required to create it. In your case of the inductor there is no field when no EMF is applied. When we apply an EMF a current flows and does work, and the work goes into creating the field. When we talk about the energy of e.g. a charge in an electrostatic field, we normally assume the charge is small enough that ...



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