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Chapter 29: Problem 68

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?

Short Answer

Expert verified
The potential difference between center and an end is 2.3 mV, between ends is 4.6 mV. These values are too small to be concerned.

Step by step solution

01

Convert Given Units

Convert the magnetic field from Gauss to Tesla and the angular speed from revolutions per minute to radians per second. The magnetic field is given as \(0.5\ \mathrm{G}\). Using the conversion \(1\ \mathrm{G} = 1\times 10^{-4}\ \mathrm{T}\), we find:\[B = 0.5\ \mathrm{G} = 0.5\times 10^{-4}\ \mathrm{T} = 5\times 10^{-5}\ \mathrm{T}\]The angular speed is \(220\ \mathrm{rpm}\). Using the conversion \(\omega = 220\ \times \frac{2\pi}{60}\), we find:\[\omega = \frac{220\times 2\pi}{60}\ \mathrm{rad/s} \approx 23.04\ \mathrm{rad/s}\]
02

Calculate Potential Difference from Center to End

The potential difference induced from the center to either end of the propeller can be calculated using the expression \[V = \frac{1}{2} B \omega L^2\].Substituting the known values, \[V = \frac{1}{2}\times 5\times 10^{-5} \times 23.04 \times (2.0)^2 \]Compute to give\[V = \frac{1}{2}\times 5\times 10^{-5} \times 23.04 \times 4 = 2.3\times 10^{-3} \mathrm{V}\].
03

Calculate Potential Difference Between the Ends

The potential difference between the two ends must consider the entire length of the propeller. For this, use the potential difference induced formula \[V = B \omega\frac{L^2}{2}\] for half the length on each side.Thus,\[V_{total} = B \omega\left(\frac{L}{2}\right)L = B \omega \frac{L^2}{2}\], the expression becomes\[V_{total} = 5\times 10^{-5}\times 23.04 \left(2.0\right) = 4.608 \times 10^{-3} \mathrm{V}\].This result of potential difference between the endpoints was found to be twice that at a single endpoint due to symmetry.
04

Analyze the Result

The final potential difference between the middle and either end is calculated as \(2.3\times 10^{-3} \mathrm{V}\), and the potential difference between the ends is \(4.608 \times 10^{-3} \mathrm{V}\). Given the context鈥攁 typical environmental magnetic field鈥攖his induced potential is very small and would not be of practical concern.

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Key Concepts

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

Magnetic Field
A magnetic field is a region in space where magnetic forces are exerted on moving charged particles. This field can be represented by field lines, where its direction is tangent to the field lines and its strength is proportional to the density of these lines. In our exercise, the magnetic field is perpendicular to the rotation plane of the propeller. The Earth's magnetic field, for example, serves as a weak magnetic field, affecting many phenomena despite its low strength of around 0.5 Gauss. Converting this to the SI unit system, we find the value to be 5 x 10^-5 Tesla, using the conversion factor of 1 Gauss = 1 x 10^-4 Tesla. Always remember: the stronger the magnetic field, the greater its potential impact on conductive and rotating objects like the propeller considered here.
Angular Velocity
Angular velocity is a measure of how quickly an object rotates or revolves around a particular point or axis. It is usually expressed in radians per second (rad/s). For practical problems around mechanical rotation, it's common to first know the speed in revolutions per minute (rpm) and then convert it into rad/s. In this solution, the propeller's speed was initially provided as 220 rpm. To find angular velocity in rad/s, the formula used is: \( \omega = \frac{220 \times 2\pi}{60} \). This results in approximately 23.04 rad/s. This conversion is crucial because many equations in physics use radians due to simplifying derivatives and integrations relating to trigonometric functions.
Potential Difference
Potential difference, also known as voltage, is the work needed per unit charge to move a charge from one point to another in an electric field. In the case of our rotating propeller within a magnetic field, a potential difference is induced due to electromagnetic induction. When calculated from the center of a rotating bar (like our propeller) to either of its ends, this potential difference can be described by the formula: \[ V = \frac{1}{2} B \omega L^2 \]where \( B \) is the magnetic field, \( \omega \) is the angular velocity, and \( L \) is the length of the bar. After finding the necessary values, this formula allows us to calculate the voltage effect at the ends, highlighting a basic principle of electromagnetism in action.
Thin Uniform Bar
A thin uniform bar is a solid object with mass and length, used in physics problems to simplify calculations by assuming even mass distribution across its length. In our scenario, the airplane propeller being modeled as a thin uniform bar means its mass and length are evenly distributed. This assumption makes it possible to apply formulas directly related to linear distribution of forces, torques, and electrical properties. When analyzing electromagnetic induction, this uniformity aids in calculating induced voltages, as the forces exerted by the magnetic field are spread consistently across the propeller's length.
Conversion of Units
Unit conversion is a fundamental skill in physics and engineering to ensure consistent and accurate calculations. In the context of this problem, we must convert magnetic field strength from Gauss to Tesla and angular speed from rpm to rad/s.
  • To convert Gauss to Tesla, use: \( 1 \text{ G} = 10^{-4} \text{ T} \). For example, 0.5 G becomes 5 x 10^-5 T.
  • To convert rpm to rad/s, multiply the rpm value by \( \frac{2\pi}{60} \) to convert from revolutions to radians and account for minutes to seconds conversion.
These conversions ensure equations yield correct results, crucial when applying physical laws to real-world phenomena.

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Most popular questions from this chapter

Displacement Current in a Wire. A long, straight, copper wire with a circular cross-scctional area of 2.1 \(\mathrm{mm}^{2}\) carries a current of 16 \(\mathrm{A}\) . The resistivity of the material is \(20 \times 10^{-8} \Omega \cdot \mathrm{m}\) . (a) What is the uniform electric field in the material? (b) If the cur- rent is changing at the rate of 4000 \(\mathrm{A} / \mathrm{s}\) , at what rate is the electric field in the material changing? (c) What is the displacement current density in the material in part (b)? (Hint: Since \(K\) for copper is very close to \(1,\) use \(\epsilon=\epsilon_{0} . )\) (d) If the current is changing as in part (b), what is the magnitude of the magnetic field 6.0 \(\mathrm{cm}\) from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of \(B\) . Is the contribution from the displacement current significant?

In a physics laboratory experiment, a coil with 200 tums enclosing an area of 12 \(\mathrm{cm}^{2}\) is rotated in 0.040 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is \(6.0 \times 0^{-5} \mathrm{T}\) . (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? b) What is the average emf induced in the coil?

A rectangle measuring 30.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) is located inside a region of a spatially uniform magnetic field of 1.25 \(\mathrm{T}\) , with the field perpendicular to the plane of the coil (Fig. 29.29 ). The coil is pulled out at a steady rate of 2.00 \(\mathrm{cm} / \mathrm{s}\) traveling perpendicular to the field lines. The region of the field ends abruptly as shown. Find the emf induced in this coil when it is (a) all inside the field: (b) partly inside the field; (c) all outside the field.

A metal ring 4.50 \(\mathrm{cm}\) in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic ficld. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 \(\mathrm{T} / \mathrm{s}\) (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

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. \((29.21) .\)
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