Self and Mutual Inductance in a Toroid

In summary, the conversation discusses calculating L1, L2 and M for coils in a toroid, as well as the induced EMF and polarity in the second coil. The second part of the conversation addresses calculating induced EMF in each coil and energy stored in the toroid, with some uncertainty about how to handle the continuous currents. The expert advises determining the energy stored in the magnetic field for each coil by calculating the work done against the induced EMF, and suggests that the mutual inductance terms may cancel out in this case.
  • #1
libelec
176
0
URGENT!: Self and mutual inductance

Homework Statement


1http://img20.imageshack.us/img20/8531/asdasdasv.png

a) Calculate L1, L2 and M for the coils in the following toroid (using thin toroid condition), the induced EMF in the second coil and the polarity.

N1 = 500
N2 = 200
I1 = (20 + 0.2 t/s)A
S = 3 cm2
rM = 5 cm
[tex]\mu[/tex]r = 1200

b) If now there's a current I2 = 2A through coil 2 from top to bottom, and I1 = 20A, calculate the induced EMF in each coil and the stored energy in the toroid.


The Attempt at a Solution



I can find all I'm asked for in a). There's no problem there. But in b) I can't see how there's going to be an induced EMF is the currents are continuous. I know that they actually rise from 0 to I2 and I1 in a period of time, but I'm not given that time. Therefor I can't use that [tex]{\varepsilon _1} = - {L_1}\frac{{d{I_1}}}{{dt}} - {M_{21}}\frac{{d{I_2}}}{{dt}}[/tex] or that [tex]{\varepsilon _2} = - {L_2}\frac{{d{I_2}}}{{dt}} - {M_{12}}\frac{{d{I_1}}}{{dt}}[/tex]. Also, because there are two different currents going through each coil, I can't think of it in terms of an equivalent inductance Leq.

What do I do?
 
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  • #2


libelec said:
I can find all I'm asked for in a). There's no problem there. But in b) I can't see how there's going to be an induced EMF is the currents are continuous.
There isn't. If dI/dt = 0 there are no induced voltage/currents. So all you have to do is determine the energy stored in the magnetic field due to the currents in each coil. This is a function of L and the current. Mutual inductance would be irrelevant.

AM
 
  • #3


So what I have to do is to calculate U as the sum of the energies stored in each coil?

Thank you.
 
  • #4


libelec said:
So what I have to do is to calculate U as the sum of the energies stored in each coil?
Think of the energy stored in the coil as the work done against the induced emf as the current builds up from 0 to the final value:

[tex]W = \int_0^{I_f} Pdt = \int_0^{I_f} \varepsilon i dt[/tex]

Write out the expression for the emf in the integral for each coil to find the energy of the magnetic field for each coil. The total energy is the sum of the two integrals.

I may have overstated it in saying that mutual inductance is irrelevant because the mutual inductance does determine the induced emf as the current builds up. But I think the two mutual inductance terms cancel each other out and you are left with just the self inductance terms. Work it out and see.

AM
 

Related to Self and Mutual Inductance in a Toroid

What is self-inductance?

Self-inductance is a property of a circuit element (such as a coil) that describes its ability to generate an induced voltage in response to a change in current flowing through it. It is measured in units of henries (H).

What is mutual inductance?

Mutual inductance is a property of two circuit elements that describes how they can induce an electromotive force (EMF) in each other. It is measured in units of henries (H).

How do self-inductance and mutual inductance affect each other?

Self-inductance and mutual inductance are closely related, as they both involve the generation of induced voltage. In a circuit with multiple elements, the presence of mutual inductance between two elements can affect the self-inductance of each element. This is because the magnetic field generated by one element can induce a voltage in the other element, which in turn affects its self-inductance.

What are some real-world applications of self and mutual inductance?

Self and mutual inductance have many practical applications, such as in transformers, electric motors, generators, and inductors. These components are commonly used in power distribution systems, electronics, and household appliances.

How can self and mutual inductance be calculated?

The equations for calculating self and mutual inductance depend on the geometry and physical properties of the circuit elements involved. For simple geometries, such as a solenoid or two parallel wires, there are specific equations that can be used. In more complex cases, numerical methods or simulation software may be necessary to calculate the values accurately.

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