Problem 62 - University Physics with Modern Physics

Question

An airplane propeller of total length \(L\) rotates around its center with angular speed \(\omega\) in a magnetic field that is perpendicular 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 G and the propeller turns at 220 rpm and is 2.0 m long, what is the potential difference between the middle and either end? It this large enough to be concerned about?

Step-by-Step Solution

Answer

The potential difference between the middle and either end is 2.30 mV. It is not large enough to be concerned about.

Step 1: Understanding Parameters

In this problem, an airplane propeller acts like a thin, uniform rod rotating in a magnetic field. Given parameters include:- Total length of the propeller, \( L \)- Angular speed of the propeller, \( \omega \)- Magnetic field, \( B \)Additionally, the earth's magnetic field is 0.50 G (which is 0.50 \times 10^{-4} \text{ T}) and the propeller turns at 220 rpm.

Step 2: Calculate the Angular Speed in SI Units

First, convert the angular speed from rotations per minute (rpm) to radians per second (rad/s).Given angular speed is 220 rpm. Convert it using the relation:\[ \omega = 220 \times \frac{2\pi}{60} \text{ rad/s} = \frac{220 \times 2\pi}{60} \text{ rad/s} \approx 23.038 \text{ rad/s} \]

Step 3: Formula for Induced EMF

The induced electromotive force (emf) due to rotation in a magnetic field is given by:\[ \varepsilon = \frac{1}{2} B \omega L^2 \]This formula gives the potential difference between the center and either end of the uniform rod (propeller).

Step 4: Calculate the Potential Difference for Part (a)

Substitute the given values into the formula for induced emf:\[ \varepsilon = \frac{1}{2} \times 0.50 \times 10^{-4} \times 23.038 \times 2^2 \]Simplify the calculation:\[ \varepsilon = \frac{1}{2} \times 0.50 \times 10^{-4} \times 23.038 \times 4 \]\[ \varepsilon = 2.30 \times 10^{-3} \text{ V} \]

Step 5: Calculate the Potential Difference for Part (b)

The potential difference between the two ends of the propeller (thin bar) is twice the potential difference between the center and one end. Therefore:\[ \varepsilon_{end-to-end} = 2 \times 2.30 \times 10^{-3} = 4.60 \times 10^{-3} \text{ V} \]

Step 6: Analyze Part (c)

The potential difference between the middle and either end is 2.30 \( \times 10^{-3} \text{ V} \). Since this voltage is very small, its effect is negligible, and it is not large enough to be concerned about.

Key Concepts

Angular Speed

Angular speed ( \( \omega \) ) is a measure of how fast an object is rotating. It's the rate at which an object rotates around a fixed point or axis. For the propeller in this exercise, angular speed is given in rotations per minute (rpm), but for calculations, it is often converted into radians per second (rad/s). This conversion is crucial as it translates the rotational movement into a standard unit compatible with other physical measurements, like those used in the formulae for electromagnetic induction.
Understanding angular speed helps determine how quickly the propeller sweeps through the magnetic field, which is essential for calculating other quantities, such as the induced electromotive force (emf). In this problem, the angular speed of the propeller is converted from 220 rpm to approximately 23.038 rad/s using the understanding that 1 rotation equals 2\( \pi \) radians and 1 minute equals 60 seconds.
This conversion is done using the formula:

• \( \omega = \frac{\text{(RPM)} \times 2\pi}{60} \)

Magnetic Field

The magnetic field ( \( B \) ) is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. It is measured in teslas (T) in the SI unit system, though it is sometimes given in gauss (G) for smaller fields. In the context of this problem, the magnetic field is perpendicular to the plane of rotation of the propeller. This orientation maximizes the interaction between the magnetic field and the rotating conductor, which in this case is the propeller modeled as a thin rod.
The Earth's magnetic field is given here as 0.50 G, which is \( 0.50 \times 10^{-4} \text{ T} \) . This may seem small, but it is sufficient to induce an electromotive force in the rotating propeller. Understanding the orientation and magnitude of the magnetic field is essential because it directly affects the calculations of the induced emf.

• Magnetic fields exert forces on charges and current carriers.
• Perpendicular alignment ensures the maximum induced emf.

Induced Electromotive Force

Induced electromotive force ( \( \varepsilon \) ), or emf, is the voltage generated by changing magnetic fields through a conductor, in this case, the rotating propeller. According to Faraday’s Law of Electromagnetic Induction, a change in magnetic flux through a loop induces an emf.
For this problem, the emf is produced because the propeller spins within a magnetic field, altering how the field lines interact with the metal of the propeller. The formula used to calculate the induced emf is:

• \( \varepsilon = \frac{1}{2} B \omega L^2 \)

Where:

• \( B \) is the magnetic field.
• \( \omega \) is the angular speed.
• \( L \) is the length of the propeller.

This formula shows that the induced emf is proportional to the square of the length of the propeller, the strength of the magnetic field, and the angular speed.

Potential Difference

Potential difference is the work needed to move a charge from one point to another in an electric field, measured in volts (V). In this scenario, the problem seeks to determine the potential difference induced by the rotation of the propeller in a magnetic field.
The calculations consider two different scenarios: between the center and either end, and between the two ends. Using the induced emf ( \( \varepsilon \) ) formula, the potential difference between the center and either end is calculated to be approximately \( 2.30 \times 10^{-3} \text{ V} \) . For the entire length of the propeller (end-to-end), this potential is doubled because each segment contributes equally to the voltage, resulting in \( 4.60 \times 10^{-3} \text{ V} \) .

• The small potential difference indicates a negligible effect, assuring us that in most situations, such small voltages on a propeller are not a cause for concern.
• Understanding potential difference is crucial as it gives insights into the electrical energy changes in the system.