Tag Info

Hot answers tagged

18

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 in ...


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 ...


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 ...


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 ...


4

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 ...


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

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

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 ...


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 ...


2

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 ...


2

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 $E$ ...


2

Your magnetic case is similar to the electric case of two opposite charges (balls) of finite density and with no polarization effects (bound charges). At infinite distance you have two equal electrostatic energies $E_1=E_2=E$ (self-energy). At a finite distance $r>2R$ you add to it the energy of electrostatic interaction $U(r)=-Q^2/r$, the latter being ...


2

Books say that ΣΔV=0 around a closed loop KVL holds only if the magnetic flux linking the circuit is unchanging. In ideal circuit theory, it is assumed that circuit elements are ideal lumped elements and the self inductance of the circuit is zero. In other words, we assume that the dimensions of the circuit and circuit elements are arbitrarily small ...


2

Reluctance = $\dfrac{l_e}{\mu A_e}$ where..... mu is the absolute permeability of the material, $\mu_0 \mu_r $ $l_e$ is the circumference of a circle at a radius r and $A_e$ is a small cross sectional area. The circle I refer to only relates to the cross section of the torus and r is the radius from the centre (where the wire is). All these reluctances ...


2

I assume the first part, up to But how exactly does it happen? defines and explains your question, and then you show what you think about it so far? It looks like the point where it goes wrong is about what the inductor does. There is nothing about "split-second" and relativistic, it behaves in a pretty symmetric way to the capacitor. It's "dynamic" ...


1

As the voltage between the capacitor's plates decreases, so should the current flowing through the circuit. I don't follow your reasoning here. Recall that, for an ideal capacitor, we have: $$i_C = C\frac{dv_C}{dt}$$ In words, the current through the capacitor is proportional to the rate of change of the voltage across, not the instantaneous value ...


1

You should write down all the equations of you LC circuit and that may help. The short answer is that when you close the switch and let current flow out of the capacitor, it can't flow right away because the rapidly changing current sets up an opposing voltage in the inductor. V = L di/dt And so the current increases slowly, reaching a maximum (as you ...


1

$\def\ddt{\frac{d}{dt}}\def\l\{\left}\def\r{\right}\def\rmS{S}$The network is over-idealized. Any inductor currents $i_1$, $i_2$ with $i_1+i_2 = \frac{e}{R}$ are possible at steady state. If you are given initial currents $i_1(0),i_2(0)$ then current over time results with $i_\rmS:=i_1+i_2$ from \begin{align} \ddt i_1(t) &= \frac1{L_1}\l(e - R\cdot ...



Only top voted, non community-wiki answers of a minimum length are eligible