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Chapter 25: Problem 2

In a physics laboratory experiment, a coil with 200 turns enclosing an area of \(12 \mathrm{~cm}^{2}\) is rotated in \(0.040 \mathrm{~s}\) 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 \(6.0 \times 10^{-5} \mathrm{~T}\). (a) What is the magnetic flux through each turn of the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?

Short Answer

Expert verified
The initial magnetic flux through each turn is \(7.2 \times 10^{-8} \, Wb \) and the final magnetic flux is \(0 \, Wb \). The average induced emf is \(1.8 \times 10^{-6} \, V \)

Step by step solution

01

Calculation of initial magnetic flux

The magnetic flux (\( \phi \)) through a coil is given by the equation \( \phi = \mathrm{BA} \cos(\alpha) \) where B is the magnetic field, A is the area, and \( \alpha \) is the angle between the normal to the coil and the magnetic field. When the coil's plane is perpendicular to the field, the angle (\( \alpha \)) is \( 0 \) degrees. Thus, the initial magnetic flux through each turn is given by \( \phi_{i} = B \times A \times \cos(0) = 6.0 \times 10^{-5} \,T \times 12 \times 10^{-4} \, m^2 = 7.2 \times 10^{-8} \, Wb \)
02

Calculation of final magnetic flux

When the coil's plane is parallel to the field, the angle (\( \alpha \)) is \( 90 \) degrees. Thus, the final magnetic flux through each turn is given by \( \phi_{f} = B \times A \times \cos(90) = 0 \, Wb \) since cosine of 90 degrees is zero.
03

Calculation of average induced emf

The average induced emf (蔚) can be calculated by the expression \( 蔚= - \frac{\Delta \phi}{ \Delta t} \) where \( 螖 \phi \) is change in flux and \( 螖 t \) is time interval. Hence \( 蔚= - \frac{\phi_{f} - \phi_{i}}{0.040 s} = \frac{ 7.2 \times 10^{-8} Wb}{0.040 s} = 1.8 \times 10^{-6} \, V \) . The negative sign indicates that the induced emf works to oppose the change in the magnetic flux (Lenz's law).

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Key Concepts

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

Magnetic Flux
Magnetic flux is a fundamental concept in electromagnetic induction. It provides a measure of the total magnetic field passing through a given area. To calculate magnetic flux (\( \phi \)), the formula is used: \( \phi = \mathrm{B \times A \times \cos(\alpha)} \). In this equation:
  • \( B \) is the magnetic field strength.
  • \( A \) is the area through which the field lines pass.
  • \( \alpha \) is the angle between the normal (a line perpendicular) to the coil and the magnetic field direction.
In the given exercise, first calculate the magnetic flux when the coil's plane is perpendicular to the magnetic field. Here, \( \alpha \) is 0 degrees, and \( \cos(0) = 1 \), simplifying the flux to \( B \times A \). Initially, the flux is \( 7.2 \times 10^{-8} \, Wb \). After the coil is rotated parallel to the field, the angle is 90 degrees, making \( \cos(90) = 0 \). This results in zero magnetic flux, as no field lines pass through the coil. Understanding these conditions helps grasp changes in magnetic flux during rotations.
Induced EMF
Induced electromotive force (EMF) is a crucial concept in understanding how rotation in a magnetic field generates electrical energy. It explains how a changing magnetic flux through a coil induces a voltage across it. The formula employed is \( \varepsilon = - \frac{\Delta \phi}{\Delta t} \), where:
  • \( \varepsilon \) represents the induced EMF.
  • \( \Delta \phi \) is the change in magnetic flux.
  • \( \Delta t \) denotes the time over which the change occurs.
In the exercise, the change in magnetic flux, \( \Delta \phi \), is from \( 7.2 \times 10^{-8} \, Wb \) to \( 0 \, Wb \), resulting in a flux change of \( -7.2 \times 10^{-8} \, Wb \) (as it's decreasing). The time period for this change is \( 0.040 \, s \). By substituting these values into the formula, we find that the average induced EMF is \( 1.8 \times 10^{-6} \, V \), with the negative sign highlighting Lenz's Law, indicating the EMF opposes the change in flux. Understanding induced EMF is key in numerous electrical applications, such as transformers and generators.
Lenz's Law
Lenz's Law provides insight into the direction of the induced EMF and current. It states that the induced EMF will always work to oppose the change in magnetic flux that produced it. This is encapsulated in the negative sign used in Faraday's law of electromagnetic induction equation. Lenz鈥檚 law is a manifestation of the law of conservation of energy and can be stated as follows:
  • If the magnetic flux through a coil increases, the induced EMF generates a current whose magnetic field opposes the increase.
  • Conversely, if the magnetic flux decreases, the induced current generates a magnetic field to oppose the decrease.
In this exercise, the initial magnetic flux decreases to zero, and the induced EMF counteracts this change by creating a current that would generate magnetic flux in the original direction, opposing the nullifying of flux. By comprehending Lenz's Law, students can better predict the behavior of induction phenomena, ensuring they grasp the safety and efficiency aspects of devices relying on electromagnetic induction, such as inductors in circuits.

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Most popular questions from this chapter

The current in a wire varies with time according to the relationship \(I=55 \mathrm{~A}-\left(0.65 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2} .\) (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{~s} ?(\mathrm{~b}) \mathrm{What}\) constant current would transport the same charge in the same time interval?

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of \(165.0 \mathrm{~cm}\), but its circumference 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, uniform 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 \(9.0 \mathrm{~s}\) have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

A material with resistivity \(\rho\) is formed into a cylinder of length \(L\) and outer radius \(r_{\text {outer }}\). A cylindrical core with radius \(r_{\text {inner }}\) is removed from the axis of this cylinder and filled with a conducting material, which is attached to a wire. The outer surface of the cylinder is coated with a conducting material and attached to another wire. (a) If the second wire has potential \(V\) greater than the first wire, in what direction does the local electric field point inside of the cylinder? (b) The magnitude of this electric field is \(c / r,\) where \(c\) is a constant and \(r\) is the distance from the axis of the cylinder. Use the relationship \(V=\int \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{l}}\) to determine the constant \(c .(\mathrm{c})\) What is the resistance of this device? (d) A \(1.00-\mathrm{cm}\) -long hollow cylindrical resistor has an inner radius of \(1.50 \mathrm{~mm}\) and an outer radius of \(3.00 \mathrm{~mm} .\) The material is a blend of powdered carbon and ceramic whose resistivity \(\rho\) may be altered by changing the amount of carbon. If this device should have a resistance of \(6.80 \mathrm{k} \Omega,\) what value of \(\rho\) should be selected?

A very long, straight solenoid with a cross-sectional area of \(2.00 \mathrm{~cm}^{2}\) is wound with 90.0 turns of wire per centimeter. Starting at \(t=0\) the current in the solenoid is increasing according to \(i(t)=\left(0.160 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2}\). A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is \(3.20 \mathrm{~A}\) ?

The battery for a certain cell phone is rated at \(3.70 \mathrm{~V}\). According to the manufacturer it can produce \(3.15 \times 10^{4} \mathrm{~J}\) of electrical energy, enough for \(5.25 \mathrm{~h}\) of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.
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