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Hot answers tagged inductance

16

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

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

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

"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 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 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

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 ...

2

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 ...

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

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 ...

2

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 . ...

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

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

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 ...

1

Here is a good start: ON THE SELF-INDUCTANCE OF CIRCLES. A quote from the text: Various formulae have been given by different authors for the self-inductance of circles; that is, for closed rings of circular cross section. Some of these formulae are at once convenient and accurate, while others are both inconvenient and unreliable, and should be ...

1

They are different because in a transmission line we have distributed resistance, capacitance, conductance and inductance (meaning that each tiny segment of transmission line has its own tiny resistance, capacitance, conductance and inductance) while in RLC circuits we have lumped resistance, inductance and capacitance. Also RLGC doesn't model a transmission ...

1

First of all, your formula for the energy density is missing a term for the electric field: $$w(t)=\frac{\varepsilon_0 e^2(t)}{2}+\frac{b^2(t)}{2\mu_0 }$$ You will always have an electric field if the magnetic field is time-varying. However, this is not that important for the more conceptional question you seem to be struggling with. I think a good ...

1

A time-varying magnetic field produces an electric field. In your example $c$ $\rightarrow$ $\infty$ -- I'll work in this limit so that we can get rid of second-order effects (i.e. the quasi-static approximation.) The relevant Maxwell equation is $$\nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}.$$ We know that the magnetic field will curl ...

1

There are a couple of nonlinear magnetic material effects that might be at play here, although this answer must be described as speculation without more detail. Both effects are more pronounced if your inductor is ungapped. 1) At very low current levels (corresponding to very low levels of magnetic field H), the inductance can be lower than nominal. (The ...

1

Your capacitance formula is not applicable here. If you can work with complex numbers, you can use the impedance of the inductor, $j \omega L$, to calculate the total impedance ($Z = v/i = R_{internal}+ j \omega L$), and then work out what L must be to yield the specified phase angle at the specified frequency.

1

I think you'll find this article (with pictures) helpful for dot convention and this one for mutual inductance. Let me know if this doesn't clear it up.

1

$V_{AB} = V_A - V_B$ so node A is more positive than node B side when $V_{AB}>0$ According to the Passive Reference Convention, current enters the positive terminal of a resistor so the current is from node A to node B. We have $V_{AB} = 2A \cdot R + 0.5A/s \cdot L = 9V$ and $V_{AB} = 2A \cdot R - 0.5A/s \cdot L = 5V$ which yields $R = 3.5 \Omega$ ...

1

Induction will work whether the pan is flat or not. However Foucault's currents (eddy currents) locally induce important temperature gradients within the metal. Unflat pan generates a non uniform heating that in turns induces large deformations or even cracks in the steel. Furthermore the power transfered by microwaves to the pan is optimal if the field ...

1

Generally those sort of passive rectifying tags need antenna which are long in one dimension. They have to a be a reasonable fraction of a wavelength - and you need relatively long wavelengths for long range at reasonable power. For example the Recco tags for ski clothing would work and are about 2" long. One issue may be the angle of them relative to your ...

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