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Are Motional emfs a Practical Source of Electricity? How fast (in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{mph}\) ) would a \(5.00-\mathrm{cm}\) copper bar have to move at right angles to a 0.650 - \(T\) magnetic field to generate 1.50 \(\mathrm{V}\) (the same as a AA battery) across its cnds? Does this scem like a practical way to generate electricity?

Short Answer

Expert verified
No, the speed required (46.15 m/s or 103.20 mph) makes this method impractical for generating electricity.

Step by step solution

01

Understanding the Formula

The formula to calculate the motional emf (Electromotive Force) is given by \( \text{emf} = B \cdot l \cdot v \), where \( B \) is the magnetic field in teslas, \( l \) is the length of the conductor in meters, and \( v \) is the velocity in meters per second. In this problem, you need to find the velocity \( v \).
02

Solving for Velocity

Rearrange the formula to solve for \( v \):\[ v = \frac{\text{emf}}{B \cdot l} \]Given: \( \text{emf} = 1.50 \text{ V} \), \( B = 0.650 \text{ T} \), and \( l = 5.00 \text{ cm} = 0.050 \text{ m} \). Substitute these values into the formula:\[ v = \frac{1.50}{0.650 \times 0.050} \]
03

Calculate Velocity in Meters per Second

Perform the calculation to find \( v \):\[ v = \frac{1.50}{0.0325} = 46.15 \text{ m/s} \]
04

Convert Velocity to Miles Per Hour

To convert from meters per second to miles per hour, use the conversion factor: \( 1 \text{ m/s} = 2.237 \text{ mph} \):\[ 46.15 \text{ m/s} \times 2.237 = 103.20 \text{ mph} \]
05

Evaluate the Practicality

A practical source of electricity would not require such a high speed. A speed of 46.15 m/s (103.20 mph) is unrealistic for a small copper bar to maintain, indicating that this method isn't practical for generating electricity.

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Key Concepts

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

Electromotive Force
Electromotive Force (EMF) is a fundamental aspect of electromagnetism, which refers to the voltage generated by a changing magnetic field or movement of a conductor in a magnetic field. It can be thought of as the energy supplied per unit charge by the source of electric current.
Understanding EMF is crucial because it is the driving force behind electric currents in circuits. The formula for calculating motional EMF is:
\( \text{emf} = B \cdot l \cdot v \)
  • \( B \) is the magnetic field strength in teslas (T).
  • \( l \) is the length of the conductor moving through the field in meters (m).
  • \( v \) is the velocity of the conductor in meters per second (m/s).
In this scenario, the challenge is to determine how fast a copper bar needs to move to produce an EMF of 1.50 volts, which is equivalent to a standard AA battery. This helps illustrate how EMF is not only theoretical but has practical implications in electricity generation.
Magnetic Field
A magnetic field is a region surrounding a magnetic material or moving electric charge within which the force of magnetism acts. In this exercise, it is measured using the tesla (T), a unit of magnetic field strength. A magnetic field of 0.650 T is used to interact with the moving conductor to generate the required EMF.
To generate an EMF, a conductor must cut through the magnetic field lines. The stronger the magnetic field, the more forcefully the conductor can influence electrical charges, thus generating a higher EMF.
Magnetic fields are invisible forces but can be visualized by lines representing the path a north magnetic pole follows. When discussing motional EMF, it’s important to think about how the movement of the conductor relative to these lines affects electric charges inside the conductor, making the concept of magnetic fields crucial to understanding the entire process.
Velocity Calculation
The calculation of velocity is crucial in determining how practical motional EMF might be as a source of electricity. Given the EMF of 1.50 volts required, the magnetic field strength of 0.650 T, and the length of the bar at 0.050 meters, we use the rearranged formula to solve for velocity:
\[ v = \frac{\text{emf}}{B \cdot l} \]
By plugging in the appropriate values:
\[ v = \frac{1.50}{0.650 \times 0.050} = 46.15 \text{ m/s} \]
This conversion shows that to generate sufficient voltage, the copper bar must move at 46.15 meters per second. The conversion to miles per hour (using the factor of 1 m/s = 2.237 mph) gives:
\[ 46.15 \text{ m/s} \times 2.237 = 103.20 \text{ mph} \]
Such a high speed for a small bar is impractical under most circumstances, and this helps illustrate the limitations of relying on motional EMF for practical electricity generation.
Practicality of Electricity Generation
Generating electricity through motional EMF, while theoretically possible, poses significant challenges in practical applications. The exercise demonstrates that to match the voltage of a simple AA battery, an unattainably high speed is required for a small copper bar.
This impractical nature indicates:
  • To achieve practical levels of electricity generation, we typically employ rotating machinery at relatively high speeds but still within feasible limits.
  • Alternatives like electromagnetic induction in large-scale power plants take advantage of high-speed rotations using turbines powered by steam, water, or wind.
The impracticality of achieving 103.20 mph with the scenario provided guides engineers and scientists to explore other more efficient and feasible methods of electricity generation, such as magnetic induction or solar panels. Understanding these limitations helps in designing systems that are more applicable in real-world scenarios.

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Most popular questions from this chapter

An airplane propeller of total length \(L\) rotates around its center with angular spced \(\omega\) in a magnctic ficld that is perpcndicular to the plane of rotation. Modeling the propeller as a thin, uniform bar, find the potential difference between (a) the center and either end of the propeller and (b) the two ends. (c) If the field is the earth's field of 0.50 \(\mathrm{G}\) and the propeller turns at 220 \(\mathrm{rpm}\) and is 2.0 \(\mathrm{m}\) long, what is the potential difference between the middle and either end? It this large enough to be concemed about?

A \(1.41-\mathrm{m}\) bar moves through a uniform, 1.20 . \(T\) magnetic field with a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) (Fig, 29.40\()\) . In cach case, find the emf induced between the ends of this bar and identify which, if any, end \((a \text { or } b)\) is at the higher potential. The bar moves in the direction of (a) the \(+x\) -axis; (b) the \(-y\) -axis; (c) the \(+z\) -axis. (d) How should this bar move so that the emf across its ends has the greatest possible value with \(b\) at a higher potential than \(a\) , and what is this maximum emf?

A long, straight solenoid with a cross-sectional area of 8.00 \(\mathrm{cm}^{2}\) is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 \(\mathrm{A}\) . A second winding of 12 tums encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average indnced emf in the second winding?

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.

Motional emfs in Transportation. Airplanes and trains move through the earth's magnetic field at rather high speeds, so it is reasonable to wonder whether this field can have a substantial effect on them. We shall use a typical value of 0.50 \(\mathrm{G}\) for the earth's field (a) The French TGV train and the Japanese "bullet train" reach speeds of up to 180 \(\mathrm{mph}\) moving on tracks about 1.5 \(\mathrm{m}\) apart. At top speed moving perpendicular to the earth's magnetic field, what potential difference is induced across the tracks as the wheels roll? Does this seem large enough to produce noticeable effects? (b) The Boeing \(747-400\) aircraft has a wingspan of 64.4 \(\mathrm{m}\) and a cruising speed of 565 \(\mathrm{mph}\) . If there is no wind blowing (so that this is also their speed relative to the ground), what is the maximum potential difference that could be induced between the opposite tips of the wings? Does this seem large enough to cause problems with the plane?

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