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A popular demonstration of eddy currents involves dropping a magnet down a long metal tube and a long glass or plastic tube. As the magnet falls through a tube, the magnetic flux changes as the magnet moves toward or away from each part of the tube. a) Which tube has the larger voltage induced in it? b) Which tube has the larger eddy currents induced in it?

Short Answer

Expert verified
Question: Compare the induced voltage and eddy currents in a metal tube and a non-metal (glass or plastic) tube when a magnet falls through them. Answer: Both the metal tube and the non-metal tube will have the same induced voltage due to the same change in magnetic flux. However, the metal tube will have larger eddy currents induced in it compared to the non-metal tube because the metal tube is a good conductor of electricity, while the non-metal tube is not.

Step by step solution

01

Understanding Faraday's Law

Faraday's Law of Electromagnetic Induction states that the induced electromotive force (EMF) or voltage in a closed loop of wire is equal to the negative rate of change of the magnetic flux passing through the loop: \( \mathrm{Induced EMF} = -\dfrac{\mathrm{d}(\mathrm{Magnetic\,Flux})}{\mathrm{d}t}\) In our case, we want to find the induced voltage in both tubes due to the change in magnetic flux as the magnet falls through them.
02

Identifying the materials of the tubes

In our case, we have a metal tube and a non-metal (glass or plastic) tube. The metal tube is a good conductor of electricity, while the glass or plastic tube is not. This fact will be important when determining the induced voltage and eddy currents.
03

Comparing induced voltage in both tubes

Since both tubes experience the same change in magnetic flux as the magnet falls, the induced voltage will also be the same according to Faraday's Law. Therefore, the answer to part (a) is that both tubes have the same induced voltage.
04

Understanding Lenz's Law and eddy currents

Lenz's Law states that the induced EMF will always create a magnetic field that opposes the change in magnetic flux that caused it. In this case, the moving magnet induces an EMF in the metal tube. The induced EMF creates a current in the tube, known as an eddy current. Eddy currents generate their own magnetic fields, which interact with the falling magnet. Since the metal tube is a good conductor, it allows these currents to flow easily. On the other hand, the glass or plastic tube does not allow eddy currents to flow easily, as it is a poor conductor. So, there will be little or no eddy currents induced in the glass or plastic tube.
05

Comparing eddy currents in both tubes

Due to the difference in conductivity, the metal tube will have larger eddy currents induced compared to the non-metal tube. The eddy currents in the metal tube will be significant, while those induced in the glass or plastic tube will be negligible. Therefore, the answer to part (b) is that the metal tube has larger eddy currents induced in it.

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Most popular questions from this chapter

A metal loop has an area of \(0.100 \mathrm{~m}^{2}\) and is placed flat on the ground. There is a uniform magnetic field pointing due west, as shown in the figure. This magnetic field initially has a magnitude of \(0.123 \mathrm{~T}\) which decreases steadily to \(0.075 \mathrm{~T}\) during a period of \(0.579 \mathrm{~s}\). Find the potential difference induced in the loop during this time.

When a magnet in an MRI is abruptly shut down, the magnet is said to be quenched. Quenching can occur in as little as \(20.0 \mathrm{~s}\). Suppose a magnet with an initial field of \(1.20 \mathrm{~T}\) is quenched in \(20.0 \mathrm{~s},\) and the final field is approximately zero. Under these conditions, what is the average induced potential difference around a conducting loop of radius \(1.00 \mathrm{~cm}\) (about the size of a wedding ring) oriented perpendicular to the field?

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The electric field near the Earth's surface has a magnitude of \(150 . \mathrm{N} / \mathrm{C}\) and the magnitude of the Earth's magnetic field near the surface is typically \(50.0 \mu \mathrm{T}\). Calculate and compare the energy densities associated with these two fields. Assume that the electric and magnetic properties of air are essentially those of a vacuum.

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