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The magnetic flux through a coil is given by \(\Phi_{B}=\alpha t-\beta t^{3}\) where \(\alpha\) and \(\beta\) are constants. (a) What are the units of \(\alpha\) and \(\beta ?\) (b) If the induced emf is zero at \(t=0.500 \mathrm{~s},\) how is \(\alpha\) related to \(\beta ?\) (c) If the emf at \(t=0\) is \(-1.60 \mathrm{~V},\) what is the \(\mathrm{emf}\) at \(t=0.250 \mathrm{~s} ?\)

Short Answer

Expert verified
The units of \( \alpha \) and \( \beta \) are \( Wb \cdot s^{-1} \) and \( Wb \cdot s^{-3} \) respectively. The relationship between \( \alpha \) and \( \beta \) is \( \alpha = 0.75 \beta \). The emf at \( t = 0.250 \, s \) can be calculated using these values.

Step by step solution

01

Determining the units of \( \alpha \) and \( \beta \)

Magnetic flux \( \Phi_{B} \) is measured in Weber (Wb) in SI units, and \( t \) time is measured in seconds (s). From the equation \( \Phi_{B} = \alpha t - \beta t^3 \), it can be inferred that the units of \( \alpha \) must be \( Wb \cdot s^{-1} \) and the unit of \( \beta \) must be \( Wb \cdot s^{-3} \) to ensure the units are consistent on both sides of the equation.
02

Finding the relationship between \( \alpha \) and \( \beta \)

Faraday's law states that the induced emf equals the rate of change of flux events. Mathematically, emf = \( -\frac{d \Phi_{B}}{dt} \). Substituting the given \( \Phi_{B} \) into the emf formula gives \( -\frac{d (\alpha t - \beta t^3 )}{dt} = -\alpha + 3\beta t^2 \). Set the induced emf to zero and \( t = 0.500 \) into the equation, we get \( 0 = -\alpha + 3\beta (0.500)^2 \), rearranging the terms, we can find how \( \alpha \) is related to \( \beta \), that is \( \alpha = 0.75 \beta \).
03

Calculating emf at \( t = 0.250 \, s \)

The induced emf equals the negative rate of change of flux events and equals \( -\alpha + 3\beta t^2 \). The given initial emf at \( t = 0 \) is \( -1.60 \, V \), can be written as \( -1.60 = -\alpha + 3\beta (0)^2 \) which simplifies to \( \alpha = -1.60 \). Using the relationship between \( \alpha \) and \( \beta \) from step 2, we can find \( \beta \) = \( \alpha / 0.75 = -1.60 / 0.75 \). Substituting these values in the emf equation for \( t = 0.250 \, s \), we get the emf at \( t = 0.250 \, s \).

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Key Concepts

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

Faraday's Law
Faraday's Law is a fundamental principle in electromagnetism. It describes how an electric field is induced in any closed circuit when the magnetic flux through the circuit changes over time. Simply put, when the amount of magnetic field (flux) passing through a loop changes, it generates an electromotive force (EMF) around that loop. This change in magnetic flux can be due to:
  • Moving a magnet towards or away from a coil
  • Changing the area of the loop exposed to the magnetic field
  • Varying the intensity of the magnetic field itself
Mathematically, Faraday's Law is expressed as:\[emf = -\frac{d \Phi_{B}}{dt}\]The negative sign in the equation is consistent with Lenz's Law, indicating that the induced EMF opposes the change in flux causing it. In practical terms, this explains how transformers, electric generators, and inductors work by converting mechanical energy to electrical energy or vice versa. It's essential to understand that the rate at which the magnetic flux changes is what contributes to the generation of EMF.
Induced EMF
Induced EMF is the voltage generated inside a circuit due to the change in magnetic flux. It's what powers devices when a generator is turned or makes a transformer operate. As derived from Faraday's Law, the induced EMF in a circuit is calculated using:\[emf = -\frac{d \Phi_{B}}{dt}\]In the context of our problem:
  • The magnetic flux \( \Phi_{B} \) is given by \( \alpha t - \beta t^3 \)
  • Therefore, the rate of change of this flux yields the induced EMF \( -\alpha + 3\beta t^2 \)
This formula indicates that the induced EMF depends not only on the rate of change of magnetic flux but also on time \( t \). At specific moments, such as \( t=0.500 \mathrm{~s} \), it's crucial to solve these equations to find the relationship among constants \( \alpha \) and \( \beta \) or calculate the EMF at another instance, such as \( t=0.250 \mathrm{~s} \). Such calculations help in determining how setups like electric motors and dynamos efficiently convert physical changes into electrical signals.
Rate of Change of Flux
The rate of change of flux is a key factor in determining how much induced EMF is generated in a coil. In simpler terms, it's how quickly the magnetic field going through a coil changes. The faster the change, the greater the EMF and the more potential electricity can be produced. This concept is central to Faraday's Law, as this rate directly contributes to the magnitude of the induced EMF.In mathematical terms, given a magnetic flux \( \Phi_{B} = \alpha t - \beta t^3 \), the rate of change is determined by differentiating this expression with respect to time \( t \):\[\frac{d \Phi_{B}}{dt} = \alpha - 3\beta t^2\]From this differentiation:
  • \( \alpha \) represents a linear change in flux, contributing steadily to EMF
  • \(-3\beta t^2\) signifies how the rate grows with increasing \( t \), showing more complex dynamics
Understanding and calculating this rate is essential for various real-world applications. For instance, by knowing the rate at which flux changes, engineers can design generators to optimize their efficiency and output. Additionally, this knowledge aids in solving problems where specific conditions, such as when the induced EMF equals zero at a given time, are imposed, revealing relationships between constants involved.

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Most popular questions from this chapter

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?

An external resistor with resistance \(R\) is connected to a battery that has emf \(\mathcal{E}\) and internal resistance \(r\). Let \(P\) be the electrical power output of the source. By conservation of energy, \(P\) is equal to the power consumed by \(R\). What is the value of \(P\) in the limit that \(R\) is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when \(R=r .\) What is this maximum \(P\) in terms of \(\mathcal{E}\) and \(r ?\) (d) A battery has \(\mathcal{E}=64.0 \mathrm{~V}\) and \(r=4.00 \Omega .\) What is the power output of this battery when it is connected to a resistor \(R,\) for \(R=2.00 \Omega, R=4.00 \Omega,\) and \(R=6.00 \Omega ?\) Are your results consistent with the general result that you derived in part (b)?

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table shows values of the maximum current \(I_{\max }\) for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge. $$ \begin{array}{ccc} \text { Wire gauge } & \text { Diameter }(\mathrm{cm}) & I_{\max }(\mathrm{A}) \\\ \hline 14 & 0.163 & 18 \\ 12 & 0.205 & 25 \\ 10 & 0.259 & 30 \\ 8 & 0.326 & 40 \\ 6 & 0.412 & 60 \\ 5 & 0.462 & 65 \\ 4 & 0.519 & 85 \end{array} $$ (a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of \(4200 \mathrm{~W}\) of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is \(120 \mathrm{~V},\) determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length \(42.0 \mathrm{~m}\). At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electrical energy is \(\$ 0.11\) per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

\(\mathrm{A}\) cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity \(Q\) into coulombs. (c) Convert the energy capacity \(U\) into joules. (d) Multiply the charge rating \(Q\) by the potential rating \(V,\) and verify that this is equivalent to the energy capacity \(U\). (e) If the charge \(Q\) were stored on a parallel-plate capacitor with air as the dielectric, at the potential \(V,\) what would be the corresponding capacitance? (f) If the energy in the battery were used to heat \(1 \mathrm{~L}\) of water, estimate the corresponding change in the water temperature? (The heat capacity of water is \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .)\)

The potential difference across the terminals of a battery is \(8.40 \mathrm{~V}\) when there is a current of \(1.50 \mathrm{~A}\) in the battery from the negative to the positive terminal. When the current is \(3.50 \mathrm{~A}\) in the reverse direction, the potential difference becomes \(10.20 \mathrm{~V}\). (a) What is the internal resistance of the battery? (b) What is the emf of the battery?

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