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Chapter 9: Problem 9

A brass disc of radius \(a\), thickness \(b\) and conductivity o has its plane perpendicular to a uniform magnetic B-field which varies according to \(B=B_{0}\) sin \(\omega t\). Assuming that eddy currents flow in concentric circles about the centre of the disc, find the total current flowing at any instant and the mean power dissipated as heat. Comment on the result as an indication of the factors affecting eddy current losses in iren.

Short Answer

Expert verified
In summary, a brass disc is placed perpendicular to a uniform magnetic field that varies with time. The total current flowing at any instant is found using Faraday's law of electromagnetic induction and Ohm's law, while the mean power dissipated as heat is calculated based on the current and resistance. The main factors affecting eddy current losses in this configuration are the size of the disc, the conductivity of the material, and the frequency of the magnetic field. In practice, engineers may use thinner sheets or laminations and materials with lower conductivity to minimize eddy current losses, as well as choosing an appropriate magnetic field frequency.

Step by step solution

01

Determine the induced emf in a single concentric circle

In order to find the induced emf, we need to use Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux through a circuit. In this case, it will be given by: \(E = -\frac{d(\Phi)}{dt}\) Where E is the induced emf and \(\Phi\) is the magnetic flux. The area of a concentric circle of radius r is given by \(A = \pi r^2\). Therefore, the magnetic flux through a single concentric circle of radius r is given by: \(\Phi = BA = B_0 \sin(\omega t) \pi r^2\) Now, we can find the induced emf by taking the derivative with respect to time: \(E = -\frac{d(\Phi)}{dt} = -\pi r^2 B_0 \omega \cos(\omega t)\)
02

Determine the resistance and induced current in a single concentric circle

With the induced emf found, we can now calculate the resistance and then the current in each concentric circle. The resistance of a circular loop with infinitesimally small thickness dr can be expressed as \(R = \frac{\rho l}{A} = \frac{2 \pi r dr}{b \sigma}\) where \(\sigma\) is the conductivity of the material. Now, we can use Ohm's law (\(E = IR\)), to find the induced current in each concentric circle: \(I = \frac{E}{R} = -\frac{\pi r^2 B_0 \omega \cos(\omega t)}{\frac{2 \pi r dr}{b \sigma}} = -\frac{b \sigma r B_0 \omega \cos(\omega t)}{2 dr}\)
03

Calculate the total current flowing through the disc

To find the total current flowing through the disc, we need to add up the currents flowing through all the concentric circles. Using the integral operation for this summation, we get the total current as: \(I_\text{total} = \int_0^a I dr = \int_0^a -\frac{b \sigma r B_0 \omega \cos(\omega t)}{2 dr} = - \frac{b \sigma B_0 \omega \cos(\omega t)}{2} \int_0^a r dr\) \(I_\text{total} = -\frac{b\sigma B_0 \omega \cos(\omega t)}{2} \left[\frac{r^2}{2}\right]_0^a = -\frac{b \sigma B_0 \omega a^2 \cos(\omega t)}{4}\)
04

Determine the mean power dissipated as heat

The power dissipated in the disc can be calculated using the equation: \(P = I^2R\) To find the mean power, we need to integrate this over one complete period of the oscillation and then divide by the period, \(T = \frac{2 \pi}{\omega}\): \(P_\text{mean} = \frac{1}{T} \int_0^T P dt = \frac{1}{T} \int_0^T I^2 R dt\) Using the induced current and resistance obtained in steps 2 and 3: \(P_\text{mean} = \frac{1}{T} \int_0^T \left(-\frac{b \sigma r B_0 \omega \cos(\omega t)}{2 dr}\right)^2 \frac{2 \pi r dr}{b \sigma} dt\) Simplifying and integrating with respect to time: \(P_\text{mean} = \frac{a^4 \sigma B_0^2 \omega^2}{32} \int_0^T \cos^2(\omega t) dt\) Using the identity \(\cos^2(\omega t) = \frac{1}{2} (1 + \cos(2\omega t))\) and then integrating: \(P_\text{mean} = \frac{a^4 \sigma B_0^2 \omega^2}{64} \left[t + \frac{\sin(2\omega t)}{4\omega}\right]_0^T\) As \(\sin(2n\pi) = 0\) for all integer values of n, we have: \(P_\text{mean} = \frac{a^4 \sigma B_0^2 \omega^2}{64} \left[\frac{2 \pi}{\omega}\right] = \frac{a^4 \sigma B_0^2 \omega}{32 \pi}\)
05

Conclusion and factors affecting eddy current losses

We've found the total current flowing through the disc at any instant to be: \(I_\text{total} = -\frac{b \sigma B_0 \omega a^2 \cos(\omega t)}{4}\) And the mean power dissipated as heat to be: \(P_\text{mean} = \frac{a^4 \sigma B_0^2 \omega}{32 \pi}\) From the formula for the mean power dissipated, we can see that the three factors affecting eddy current losses in such a configuration are: 1. The size of the disc (\(a\)): As the radius of the disc increases, the eddy current losses increase, which can be proportional to the fourth power of the radius. 2. The conductivity of the material (\(\sigma\)): The higher the conductivity, the greater the eddy current losses will be. 3. The frequency of the magnetic field (\(\omega\)): The eddy current losses will also increase with the frequency of the magnetic field. In the context of iron, to minimize eddy current losses, engineers might use thinner sheets or laminations and reduce the conductivity by using steel with a higher silicon content, as well as choosing an appropriate magnetic field frequency.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law of Electromagnetic Induction
The Faraday's law of electromagnetic induction is a fundamental principle that explains how electric currents are generated by changing magnetic fields. At its core, this law states that when the magnetic flux through a conductor changes, an electromotive force (EMF) is induced in the conductor. This phenomenon is the cornerstone of many electrical devices, including transformers and generators.

The induced EMF (\(E\)) can be calculated by taking the negative rate of change of the magnetic flux (\(\frac{d(\text{\Phi})}{dt}\)) through the material. Mathematically, Faraday's law is expressed as:
\[E = -\frac{d(\text{\Phi})}{dt}\]
When we apply Faraday's law to the exercise, we see the oscillating magnetic field inducing EMF in concentric circles within the brass disc. To understand eddy currents and their effects, it's vital to grasp this relationship between the changing magnetic flux and the generated current.
Electromagnetic Power Dissipation
Electromagnetic power dissipation refers to the conversion of electrical power into heat due to the resistance within a conductor. When electric current flows through a material, some energy is inevitably lost due to the material's inherent resistance. This loss is observed as heat, which is why, for instance, electrical components become warm or even hot to the touch when in use.

In our example of the brass disc within a changing magnetic field, the induced eddy currents lose energy due to the resistance of the material. The power dissipated (\(P\b)) as heat for any given moment can be calculated by the formula:
\[P = I^2R\]
Mean power dissipation (\(P_\text{mean}\)) is particularly important when considering the thermal effects of fluctuating currents in devices. It is the average power dissipated over a complete cycle of the current's oscillation. By integrating the power dissipation over time, we can determine the mean power dissipation, which helps us judge how much energy is lost in a device—as observed in the exercise solution.
Magnetic Flux
Magnetic flux is a measurement of the total magnetic field that passes through a given area. It's essentially a way to quantify the strength of a magnetic field over a specified surface. The mathematical representation of magnetic flux (\(\Phi\)) through a surface with area (\(A\)) in the presence of a magnetic field (\(B\)) is given by the formula:
\[\Phi = BA\]
In the context of our brass disc experiment, the area considered is that of a concentric circle within the disc, and the magnetic field is sinusoidally varying with time, as expressed by \(B=B_0\text{sin}(\omega t)\). This leads to a magnetic flux that is also time-dependent, which in effect produces the eddy currents due to electromagnetic induction.

The magnetic flux concept is essential in understanding how different factors, such as the area of the material through which the field is passing and the strength of the magnetic field itself, influence the induction process. These relationships highlight the impact of magnetic flux on the magnitude of induced currents and the consequent power dissipation.

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Most popular questions from this chapter

Estimate the order of magnitude of the self-inductance of an air-cored solenoid of length \(20 \mathrm{~cm}\) with one layer of 10 turns per \(\mathrm{cm}\) cach turn forming a circle of radius \(2 \mathrm{~cm}\).

Two coils \(\mathrm{A}\) and \(\mathrm{B}\) have self-inductances \(L_{1}\) and \(L_{2}\). When a steady current flows in \(A\), a quarter of its magnetic flux links \(B\). Find the proportion of the flux from a current in \(\mathrm{B}\) which links \(\mathrm{A}\). Evaluate the coefficient of coupling.

In the simple a.c. generator in which the electromotance is given by (9.9) show that the torque opposing the motion is \(\phi_{m} I\) sin \(\omega t\), where \(I\) is the armature current and \(\phi_{m}\) is the maximum flux linkage with the coil.

The self-inductance of a coil is \(20 \mathrm{mH}\). When a current flows in it, one-fifth of its magnetic flux links a second coil. What electromotance is induced in the second coil if a current in the first collapses at a uniform rate of \(0.5 \mathrm{~A} \mathrm{~s}^{-1} ?\)

Show how Lenz's law can be used to find the direction of induced current flow in (a) a circular coil rotating about a diameter in a uniform magnetic field and (b) a stationary circular coil towards which a coaxial current-carrying coil is moving.
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