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Chapter 29: Problem 52

Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could uperate using a generalor but you have nu maguels. The eardis magnetic field at your location is horizontal and has magnitude 8.0 \(\times 10^{-5} \mathrm{T}\) , and you decide to try to use this field for a generator by rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 \(\mathrm{V}\) and estimate that you can rotate the coil at 30 \(\mathrm{rpm}\) by turning a crank handle. You also decide that to have an acceptable coil resistance, the maximum mumber of turns the coil can have is 2000 . (a) What area must the coil have? (b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates? Do you this device is feasible? Explain.

Short Answer

Expert verified
The coil must have an area of 17.9 m虏. Max translational speed: 7.51 m/s. It's not feasible due to size and effort.

Step by step solution

01

Understand the Required Equation

To find the area, we need to use Faraday's Law of Electromagnetic Induction. The expression for the peak emf induced in a rotating coil is given by:\[ \text{EMF} = NBA\omega \cos(\theta) \] where N is the number of turns, B is the magnetic field, A is the area of the coil, and \(\omega\) is the angular velocity of the coil. \(\theta\) is the angle between the plane of the coil and the magnetic field, which is 90 degrees for maximum EMF, making \(\cos(\theta) = 1\).
02

Calculate Angular Velocity

The angular velocity (\(\omega\)) can be calculated using the relation:\[ \omega = \frac{2\pi \times \text{revolutions per minute}}{60} \] Given that the coil rotates at 30 rpm, \(\omega = \frac{2\pi \times 30}{60} = \pi \) radians per second.
03

Rearrange the Equation to Solve for Area

Rearrange the expression for the peak EMF to solve for the area (A):\[ A = \frac{\text{EMF}}{NB\omega} \]Substitute the known values: EMF = 9.0 V, N = 2000, B = 8.0 \times 10^{-5} T, \(\omega = \pi \) rad/s.
04

Substitute the Values and Calculate Area

Substitute the given values into the equation to calculate the area:\[ A = \frac{9.0}{2000 \times 8.0 \times 10^{-5} \times \pi} \approx 17.9 \text{ m}^2 \].
05

Determine Maximum Translational Speed

The maximum translational speed (v) of a point on the coil is given by \( v = r \cdot \omega \), where r is the radius of the coil. If the coil is circular, the radius r is derived from the area using \( \pi r^2 = 17.9 \). Hence,\[ r = \sqrt{\frac{17.9}{\pi}} \approx 2.39 \text{ m}\].Then, the maximum translational speed is:\[ v = 2.39 \cdot \pi \approx 7.51 \text{ m/s} \].
06

Evaluate Feasibility

Given the large required area and the resulting maximum speed, constructing such a coil by spinning around a crank seems impractical due to its bulk and the physical effort required to maintain rotation. Thus, the idea, while theoretically possible, might not be feasible in a practical scenario.

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Key Concepts

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

Electromagnetic Induction
Electromagnetic induction is a fundamental concept in physics that describes how a changing magnetic field induces an electromotive force (emf) in a conductor. The principle is best captured by Faraday's Law, which states that the induced emf is directly proportional to the rate of change of the magnetic flux linked with the circuit. In the context of the generator problem from the textbook exercise, this involves rotating a coil within a magnetic field, which is achieved by the Earth's horizontal magnetic field on the island.
  • The key factor here is the change in the magnetic field experienced by the coil as it rotates.
  • Without a magnet, utilizing Earth's magnetic field for generation becomes a unique challenge.
Understanding this concept helps explain why different coil dynamics (like speed and number of turns) affect the overall emf produced.
Angular Velocity
Angular velocity is an essential aspect to understand how fast an object rotates around an axis. It is particularly vital when dealing with rotating systems like generators. In this exercise, the generator's efficiency relies significantly on how fast the coil is rotated.
To find the angular velocity (\( \omega \)), the formula \[ \omega = \frac{2\pi \times \text{revolutions per minute}}{60} \]is used. Here, it translates the rotation speed from revolutions per minute (rpm) to radians per second, which is a standard measurement in physics.
  • For the coil rotating at 30 rpm, the angular velocity is calculated as \( \pi \) radians per second.
  • This measure allows us to further calculate the generated emf and the translational speed of the coil.
EMF Calculation
Calculating the electromagnetic force (emf) is the core challenge in this exercise, as it requires balancing multiple factors to reach the desired peak emf of 9.0V. The formula from Faraday's Law \(\text{EMF} = NBA\omega \cos(\theta)\)helps calculate this. Here,\( N \) is the number of turns, \( B \) is the magnetic field strength, \(\omega\) is the angular velocity, and \(\cos(\theta)\) accounts for the orientation of the coil relative to the magnetic field, which is 1 for the maximum since \(\theta = 90^\circ\).
To find the area (\( A \)), the equation is rearranged: \[ A = \frac{\text{EMF}}{NB\omega} \].By substituting the known values and solving, we determine the coil must have an area of approximately 17.9 \( m^2 \) to achieve the desired emf.
Circular Coil Design
The design of the coil greatly affects the functionality of the generator. A circular coil is practical but requires careful consideration of size. From the calculated area of 17.9 \( m^2 \), the radius, \( r \), is determined using the area formula for a circle, \(\pi r^2\). Solving for \( r \) gives approximately 2.39 meters.
This large radius impacts the generator's operation:
  • The maximum translational speed of a point on the rotating coil is calculated as \( v = r \cdot \omega \), approximately 7.51 \( m/s \), highlighting potential feasibility issues.
  • Such a large, fast-spinning coil may not be easily managed manually on a deserted island, rendering the setup impractical even if theoretically sound.

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Most popular questions from this chapter

emf in a Bullet. At the equator, the earth's magnetic field is approximately horizontal, is directed towand the north, and has a value of \(8 \times 10^{-5} \mathrm{T}\) . (a) Estimate the emf induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot toward the east. Assume the bullet has a length of 1 \(\mathrm{cm}\) and a diamcter of 0.4 \(\mathrm{cm}\) and is traveling at 300 \(\mathrm{m} / \mathrm{s}\) . Which is at higher potential: the top or boutom of the bullet? (b) What is the emfif the bullet travels south?(c) What is the emf induced between the front and back of the bullet for any horizontal velocity?

Shrinking Loop. A circular loop of flexible iron wire has an initial circunference of \(165.0 \mathrm{cm},\) but its circunference is decreasing at a constant rate of 12.0 \(\mathrm{cm} / \mathrm{s}\) due to a tangential pull on the wire. The loop is in a constant, viniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 \(\mathrm{T}\) . (a) Find the emf induced in the loop at the instant when 9.0 s have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

It is impossible to have a uniform electric field that abruptly drops to zero in a region of space in which the magnetic field is constant and in which there are no electric charges. To prove this statement, use the method of contradiction: Assume that such a case is possible and then show that your assumption contradicts a law of nature. (a) In the bottom half of a piece of paper, draw evenly spaced horizontal lines representing a uniform electric field to your right. Use dashed lines to draw a rectangle abcda with horizontal side ab in the electric-field region and horizontal side \(c d\) in the top half of your paper where \(E=0 .\) (b) Show that integration around your rectangle contradicts Faraday's law, Eq. \((29.21) .\)

A coil 4.00 \(\mathrm{cm}\) in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to \(B=(0.0120 \mathrm{T} / \mathrm{s}) t+\left(3.00 \times 10^{-5} \mathrm{T} / \mathrm{s}^{4}\right) t^{4} .\) The coil is connected to a \(600-\Omega\) resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time \(t=5.00 \mathrm{s} ?\)

The magnetic field within a long, straight solenoid with a eireular cross section and radius \(R\) is increasing at a rate of \(d B / d t .\) (a) What is the rate of change of flux through a circle with radius \(r_{1}\) inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance \(r_{1}\) from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance \(r_{2}\) from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance \(r\) from the axis from \(r=0\) to \(r=2 R\) . (e) What is the magnitude of the induced emf in circular turn of radius \(R / 2\) that has its center on the solenoid axis? (f) What is the magnitude of the indnced emf if the radins in part (e) is \(R ?(g)\) What is the induced emf if the radius in part \((e)\) is 2\(R ?\)
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