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Chapter 7: Problem 23

Vibrating wire \(*\) A taut wire passes through the gap of a small magnet (Fig. 7.33), where the field strength is 5000 gauss. The length of wire within the gap is \(1.8 \mathrm{~cm}\). Calculate the amplitude of the induced alternating voltage when the wire is vibrating at its fundamental frequency of \(2000 \mathrm{~Hz}\) with an amplitude of \(0.03 \mathrm{~cm}\), transverse to the magnetic field.

Short Answer

Expert verified
The maximum induced voltage is 3.40 mV.

Step by step solution

01

Calculate the Maximum Speed of the Wire

Firstly, the maximum speed of the wire needs to be calculated. The formula to do this is: \(v_{max}= 2 \times \pi \times f \times A\), where \(f=2000 Hz\) is the frequency and \(A=0.03 cm\) is the amplitude. Substituting these values into the formula gives us \(v_{max} = 2 \times \pi \times 2000 Hz \times 0.03 cm = 376.99 cm/s.\)
02

Calculate the Maximum Induced Voltage

Next, the maximum induced voltage can be calculated. The formula to do this according to the Faraday's law is: \(V_{max}=Blv_{max}\), where \(B=5000 Gauss\) is the magnetic field (which needs to be converted to Tesla by multiplying by \(10^{-4} T/Gauss\)), \(l=1.8 cm\) is the length of the wire in the field (which should be converted to meters by multiplying by \(0.01 m/cm\)), and \(v_{max}=376.99 cm/s\) is the maximum speed calculated in the previous step (which needs to be converted to m/s by multiplying by \(0.01 m/cm\)). Substituting these values into the formula gives us \(V_{max} = 5000 Gauss \times 10^{-4} T/Gauss \times 1.8 cm \times 0.01 m/cm \times 376.99 cm/s \times 0.01 m/cm = 3.40 \times 10^{-3} V = 3.40 mV.\)

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Key Concepts

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

Vibrating Wire
Imagine a wire tightly stretched so it cuts through the magnetic field of a small magnet. This is a common scenario in physics used to explore the principles of electromagnetism. As the wire vibrates, it moves back and forth through the magnetic field. This dynamic creates interesting electrical properties. The wire's vibration is essential because it is the primary source of induced voltage due to the changing magnetic flux. Think of the wire like a tiny guitar string oscillating in time to an external force.
Magnetic Field Strength
The magnetic field strength, denoted as "B," is crucial in calculating the induced voltage when the wire vibrates. In this exercise, the magnetic field strength is given as 5000 Gauss. To understand this better, remember that Gauss is a unit of magnetic flux density. In practice, we often convert this to Tesla for calculations, where 1 Tesla equals 10,000 Gauss. Thus, 5000 Gauss converts to 0.5 Tesla. A strong magnetic field is key to inducing a higher voltage. The strength of the field tells us how many magnetic force lines pass through an area. The greater the field strength, the greater the effect on moving charges.
Induced Voltage
Induced voltage arises from a wire's interaction with a magnetic field. According to Faraday's Law of Induction, a change in magnetic environment of a coil or conductor generates electromotive force (EMF). When the vibrating wire oscillates through the magnetic field, the length of wire and speed of vibration determine the output of induced voltage. Faraday's Law tells us that induced voltage depends on three main factors:
  • The speed of movement through the magnetic field or the wire's velocity.
  • The strength of the magnetic field, providing ample force for induction.
  • The length of wire present in the magnetic field.
This exercise shows that these factors combine to produce an induced voltage of about 3.40 mV holding everything constant.
Fundamental Frequency
The fundamental frequency of vibration marks the wire's base oscillation rate. For this wire, it's described as vibrating at 2000 Hz. To visualize this concept, imagine plucking a single guitar string and hearing the deep note that follows. This note is essentially the fundamental frequency. In our scenario, the wire achieves its maximum speed at this frequency when it vibrates transversely in the magnetic field. The speed of these vibrations influences the induced voltage critically, ensuring efficient flux change proportional to the variations in frequency.
Amplitude
Amplitude refers to the maximum extent of a vibration or movement from the central position where the wire is at rest. Measured as a distance, in this exercise, it's 0.03 cm. Amplitude plays a big role because it directly affects how much the wire moves through the magnetic field. As the wire goes up to its maximum speed, the amplitude becomes a vital part of the speed equation. In fact, higher amplitudes mean greater velocities, which translates to increased induced voltage, following Faraday's principle. Amplitude here signifies how energetically the wire vibrates, directly impacting the electrical output.

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Most popular questions from this chapter

Critical frequency of a dynamo *** A dynamo like the one in Exercise \(7.47\) has a certain critical speed \(\omega_{0}\). If the disk revolves with an angular velocity less than \(\omega_{0}\), nothng happens. Only when that speed is attained is the induced \(\mathcal{E}\) arge enough to make the current large enough to make the mag- netic field large enough to induce an \(\mathcal{E}\) of that magnitude. The critial speed can depend only on the size and shape of the conductors, ic dimension expressing the size of the dynamo, such as the radius f the disk in our example. a) Show by a dimensional argument that \(\omega_{0}\) must be given by a relation of this form: \(\omega_{0}=K / \mu_{0} \sigma d^{2}\), where \(K\) is some dimensionless numerical factor that depends only on the arrangement and relative size of the various parts of the dynamo. (b) Demonstrate this result again by using physical reasoning that relates the various quantities in the problem \((R, \mathcal{E}, E, I, B,\), etc.). You can ignore all numerical factors in your calculations and absorb them into the constant \(K\). Additional comments: for a dynamo of modest size made owever, with \(d\) measured in hundreds of kilometers rather than neters, the critical speed is very much smaller. The earth's magnetic field is almost certainly produced by a nonferromagnetic dynamo involving motions in the fluid metallic core. That fluid happens to be molten iron, but it is not even slightly ferromagnetic because it is too hot. (That will be explained in Chapter 11.) We don't know how the conducting fluid moves, or what configuration of electric currents and magnetic fields its motion generates in the core. The magnetic field we observe at the earth's surface is the external field of the dynamo in the core. The direction of the earth's field a million years ago is preserved in the magnetization of rocks that solidified at that time. That magnetic record shows that the field has reversed its direction nearly 200 times in the last 100 million years. Although a reversal cannot have been instantaneous (see Exercise 7.46), it was a relatively sudden event on the geological time scale. The immense value of paleomagnetism as an indelible record of our planet's history is well explained in Chapter 18 of Press and Siever (1978).

\(M\) for two rings ** Derive an approximate formula for the mutual inductance of two circular rings of the same radius \(a\), arranged like wheels on the same axle with their centers a distance \(b\) apart. Use an approximation good for \(b \gg a\).

Decay time for current in the earth \(* *\) Magnetic fields inside good conductors cannot change quickly. We found that current in a simple inductive circuit decays exponentially with characteristic time \(L / R\); see Eq. (7.71). In a large conducting body such as the metallic core of the earth, the "circuit" is not easy to identify. Nevertheless, we can find the order of magnitude of the decay time, and what it depends on, by making some reasonable approximations. Consider a solid doughnut of square cross section, as shown in Fig. 7.42, made of material with conductivity \(\sigma\). A current \(I\) flows around it. Of course, \(I\) is spread out in some manner over the cross section, but we shall assume the resistance is that of a wire of area \(a^{2}\) and length \(\pi a\), that is, \(R \approx \pi / a \sigma .\) For the field \(B\) we adopt the field at the center of a ring with current \(I\) and radius \(a / 2\). For the stored energy \(U\), a reasonable estimate would be \(B^{2} / 2 \mu_{0}\) times the volume of the doughnut. Since \(d U / d t=-I^{2} R\), the decay time of the energy \(U\) will be \(\tau \approx U / I^{2} R\). Show that, except for some numerical factor depending on our various approximations, \(\tau \approx \mu_{0} a^{2} \sigma .\) The radius of the earth's core is \(3000 \mathrm{~km}\), and its conductivity is believed to be \(10^{6}(\mathrm{ohm}-\mathrm{m})^{-1}\), roughly one-tenth that of iron at room temperature. Evaluate \(\tau\) in centuries.

Mutual-inductance symmetry \(* *\) In Section \(7.7\) we made use of the vector potential to prove that \(M_{12}=M_{21} .\) We can give a second proof, this time in the spirit of Exercise 3.64. Imagine increasing the currents in two circuits gradually from zero to the final values of \(I_{1 \mathrm{f}}\) and \(I_{2 \mathrm{f}}(\) "'f" for "final"). Due to the induced emfs, some external agency has to supply power to increase (or maintain) the currents. The final currents can be brought about in many different ways. Two possible ways are of particular interest. (a) Keep \(I_{2}\) at zero while raising \(I_{1}\) gradually from zero to \(I_{1 \mathrm{f}}\). Then raise \(I_{2}\) from zero to \(I_{2 \mathrm{f}}\) while holding \(I_{1}\) constant at \(I_{1 \mathrm{f}}\). (b) Carry out a similar program with the roles of 1 and 2 exchanged, that is, raise \(I_{2}\) from zero to \(I_{2} \mathrm{f}\) first, and so on.

Total charge ** A circular coil of wire, with \(N\) turns of radius \(a\), is located in the field of an electromagnet. The magnetic field is perpendicular to the coil (that is, parallel to the axis of the coil), and its strength has the constant value \(B_{0}\) over that area. The coil is connected by a pair of twisted leads to an external resistance. The total resistance of this closed circuit, including that of the coil itself, is \(R\). Suppose the electromagnet is turned off, its field dropping more or less rapidly to zero. The induced electromotive force causes current to flow around the circuit. Derive a formula for the total charge \(Q=\int I d t\) that passes through the resistor, and explain why it does not depend on the rapidity with which the field drops to zero.
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