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20

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 ...


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 ...


7

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 an 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

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

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


4

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 ...


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

Yes you're correct . We willingly don't consider it , it is perfectly correct . However , like in a DC circuit , we think current is established instantaneously . It isn't , first there is a very Induced EMF , and no current flows in circuit . and then there's falling of the EMF with time and it is only at$\infty$ time , that we get the so called current . ...


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

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

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

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

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

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 ...


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

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 ...


3

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, ...


2

Essentially, the answer to your question is yes but your equation is not quite in the general form. Typically, impedance is $$Z=R + jX$$ with $R$ being the resistance, and $X$ being the reactance which is almost the equation you show, but without the imaginary component. Specifically, $$X = \omega L - \frac{1}{\omega C}$$. What this means is that a ...


2

Expanding on Jan Dvorak's comment: When you change the magnetic field inside a loop, an emf (electromotive force) will be generated. Now if you have two loops, each of these will experience the same e.m.f. When you put them in series, you have a coil with two loops, or two coils with one loop. No matter which way you look at it the voltage across them ...


2

The flaw in your reasoning is that the potential difference (voltage) across capacitors and inductors are equal. It occurs to me that the text is probably describing the case of a parallel circuit, whereby such an equality is indeed true. However, you are analysing a series circuit, for which the constant physical quantity is current and the sum of the ...


2

What you describe essentially happens in superconductors. There "perpetual electricity" exists in the form of super currents. In normal inductors, though, the resistance of the conductor steadily dissipates the current leading to field collapse.


2

I already got the answer. because transformers simply doesn't work in high voltage environments. High voltage environments causes dielectric breakdown which reduce the transformers function which is bad. Tesla coils fix this problem.


2

There is a very intuitive way to understand why this must be so. The ideal transformer does not dissipate energy; there is no energy loss (and certainly no energy gain). Thus, if power is delivered (by an external circuit) to the primary, it must be supplied (to another external circuit) by the secondary. If follows that if the primary and secondary ...



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