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Two \(0.68-\mathrm{m}\) -long conducting rods are rotating at the same speed in opposite directions, and both are perpendicular to a 4.7-T magnetic field. As the drawing shows, the ends of these rods come to within \(1.0 \mathrm{mm}\) of each other as they rotate. Moreover, the fixed ends about which the rods are rotating are connected by a wire, so these ends are at the same electric potential. If a potential difference of \(4.5 \times 10^{3} \mathrm{V}\) is required to cause a 1.0 -mm spark in air, what is the angular speed (in \(\mathrm{rad} / \mathrm{s}\) ) of the rods when a spark jumps across the gap?

Short Answer

Expert verified
The angular speed is approximately 4160 rad/s.

Step by step solution

01

Understanding the Problem

We have two conducting rods rotating in opposite directions within a magnetic field. A potential difference is needed for a spark to occur. We are to find the angular speed at which the potential difference due to rotation equals the required potential to spark.
02

Applying Faraday's Law of Induction

The induced voltage (or electromotive force, EMF) in a rotating rod in a magnetic field can be represented by the formula: \( V = \frac{1}{2} B \omega L^2 \), where \( V \) is the potential difference, \( B \) is the magnetic field strength, \( \omega \) is the angular speed, and \( L \) is the length of the rod.
03

Setting the Known Values

We know that \( V = 4500 \, \text{V} \), \( B = 4.7 \, \text{T} \), and \( L = 0.68 \, \text{m} \). We need to solve for \( \omega \). The challenge is to find \( \omega \) when \( V \) caused by the rotation of the rods equals the potential required for a spark.
04

Solving the Equation for Angular Speed

Substitute the values into the formula: \( 4500 = \frac{1}{2} \times 4.7 \times \omega \times (0.68)^2 \). First, calculate \( (0.68)^2 = 0.4624 \). Substitute to get: \( 4500 = \frac{1}{2} \times 4.7 \times \omega \times 0.4624 \). Multiply and simplify: \( 4500 = 1.08276 \times \omega \). Solve for \( \omega \): \( \omega = \frac{4500}{1.08276} \approx 4156.97 \).
05

Concluding the Result

Thus, the angular speed \( \omega \) required for a spark to jump across the gap is approximately \( 4160 \, \text{rad/s} \) when rounded to three significant figures.

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Key Concepts

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

Faraday's Law of Induction
Faraday's Law of Induction is a fundamental principle that describes how electric currents can be induced by changing magnetic fields. In simple terms, when you have a loop of wire and the magnetic field inside the loop changes, an electric current will flow in the wire. This concept is extremely important in understanding how many types of electrical equipment operate, from transformers to electric generators.

For a rotating rod in a magnetic field, like the one in our problem, Faraday's law helps us calculate the induced voltage, or electromotive force (EMF), generated. In this case, the induced EMF is calculated using the formula: \( V = \frac{1}{2} B \omega L^2 \).
  • \( V \): The electromotive force or potential difference
  • \( B \): The magnetic field strength
  • \( \omega \): The angular speed of the rod
  • \( L \): The length of the rod
This equation shows that the induced voltage depends on three factors: the strength of the magnetic field, the speed at which the rod is rotating, and the square of the length of the rod. Understanding how these factors interact helps us determine how to achieve a particular voltage, which is crucial in real-world applications, such as generating electricity or in this exercise's context, achieving a potential difference sufficient to spark.
Angular Speed Calculation
Angular speed is a measure of how fast an object rotates or spins. In physics, it's a key concept for systems involving rotational motion. The unit used for angular speed is radians per second (rad/s), which is different from linear speed measured in meters per second. In our exercise, angular speed is crucial as it directly impacts the potential difference created by the rotating rods.

To find the angular speed needed to produce a specific voltage, we use the modified Faraday's law formula: \( V = \frac{1}{2} B \omega L^2 \). Rearranging this to solve for angular speed \( \omega \), we have:
\[ \omega = \frac{2V}{B L^2} \] By substituting the known values for the potential difference (4500 V), magnetic field strength (4.7 T), and length of the rod (0.68 m), we can compute \( \omega \). Calculating each value carefully and ensuring units are consistent let us determine that the angular speed required is approximately 4160 rad/s.

Mastering angular speed calculations not only solve textbook problems but also plays an essential role in engineering and physics, allowing us to design components in engines, turbines, and other devices that rely on rotational motion.
Potential Difference
Potential difference, often referred to as voltage, is a measure of the electrical potential energy between two points in a circuit. It's what "pushes" the electric charge to flow through a conductor, such as a wire or a rod. In this exercise, the required potential difference is what allows a spark to jump across the small gap between the ends of two rotating rods.

The potential difference needed to cause a spark depends on several factors, such as air resistance or the distance between charges. In air, it typically requires a minimum potential difference to initiate a spark. For our problem, the necessary potential difference is given as 4500 V, which is critical for the spark to occur.

By applying Faraday's Law, we calculated the potential difference generated by the rotating rods and set it equal to the required 4500 V to compute the correct angular speed for the spark. Understanding potential difference is vital in fields like electronics, as it enables safe and efficient design of circuits capable of handling specified amounts of power without causing unintended sparks or electric arcs.

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Most popular questions from this chapter

A uniform magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of \(0.80 \mathrm{V}\) and a current of \(3.2 \mathrm{A}\) are induced in the coil. The wire is then re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?

In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil \(1,\) which has 184 loops, is \(2.82 \mathrm{V}\). The emf induced in coil 2 is 4.23 V. How many loops does coil 2 have?

A flat circular coil with 105 turns, a radius of \(4.00 \times 10^{-2} \mathrm{m}\), and a resistance of \(0.480 \Omega\) is exposed to an external magnetic field that is directed perpendicular to the plane of the coil. The magnitude of the external magnetic field is changing at a rate of \(\Delta B / \Delta t=0.783 \mathrm{T} / \mathrm{s}\), thereby inducing a current in the coil. Find the magnitude of the magnetic field at the center of the coil that is produced by the induced current.

Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume that your house is the only one using electric power. At a substation the primary coil of a step-down transformer (turns ratio \(=1: 29\) ) receives the voltage from the high-voltage transmission line. Because of your usage, a current of \(48 \mathrm{mA}\) exists in the primary coil of this transformer. The secondary coil is connected to the primary of another step-down transformer (turns ratio \(=1: 32\) ) somewhere near your house, perhaps up on a telephone pole. The secondary coil of this transformer delivers a \(240-\mathrm{V}\) emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.

In some places, insect "zappers," with their blue lights, are a familiar sight on a summer's night. These devices use a high voltage to electrocute insects. One such device uses an ac voltage of \(4320 \mathrm{V},\) which is obtained from a standard \(120.0-\mathrm{V}\) outlet by means of a transformer. If the primary coil has 21 turns, how many turns are in the secondary coil?

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