Problem 23 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \( v \).

Step 2: 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:\[ v = \frac{1.50}{0.650 \times 0.050} \]

Step 3: Calculate Velocity in Meters per Second

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

Step 4: Convert Velocity to Miles Per Hour

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

Step 5: 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.

Key Concepts

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.