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Chapter 19: Problem 45

A generator has a 150 -turn square coil, \(5 \mathrm{~cm}\) on a side. Find the peak emf in this coil when it's rotating at \(100 \mathrm{~Hz}\) (a) in Earth's field, \(B=5 \times 10^{-5} \mathrm{~T},\) and \((\mathrm{b})\) in a strong \(6.4-\mathrm{T}\) field.

Short Answer

Expert verified
(a) \( \mathcal{E}_{\text{peak}} = 0.118 \, \text{V} \), (b) \( \mathcal{E}_{\text{peak}} = 15066 \, \text{V} \).

Step by step solution

01

Identify Given Values

We have a generator with a 150-turn square coil with each side measuring 5 cm, rotating at a frequency of 100 Hz. For part (a), the magnetic field is Earth鈥檚 field: \(B = 5 \times 10^{-5} \text{ T}\). For part (b), the magnetic field is \(B = 6.4 \text{ T}\).
02

Calculate Coil Area

The area of the coil (A) can be found using the side length of the square: \(A = (0.05 \, \text{m})^2 = 0.0025 \, \text{m}^2\).
03

Determine Angular Velocity

The angular velocity (\(\omega\)) is calculated using the frequency: \(\omega = 2\pi f = 2\pi \times 100 \, \text{rad/s}\).
04

Calculate Peak EMF in Earth's Field

For part (a), use Faraday's law of induction to calculate the peak EMF: \( \mathcal{E}_{\text{peak}} = NAB\omega \), where \(N = 150\), \(A = 0.0025 \, \text{m}^2\), \(B = 5 \times 10^{-5} \, \text{T}\), and \(\omega = 2\pi \times 100 \, \text{rad/s}\). Substituting, we get: \( \mathcal{E}_{\text{peak}} = 150 \times 0.0025 \times 5 \times 10^{-5} \times 2 \pi \times 100\). Calculate this to find \(\mathcal{E}_{\text{peak}}\).
05

Calculate Peak EMF in Strong Field

For part (b), use the same formula: \( \mathcal{E}_{\text{peak}} = NAB\omega \), where \(B = 6.4 \text{ T}\). Substitute the known values: \( \mathcal{E}_{\text{peak}} = 150 \times 0.0025 \times 6.4 \times 2 \pi \times 100\). Calculate this to find \(\mathcal{E}_{\text{peak}}\).
06

Complete the Calculations

For part (a), the peak EMF is calculated as: \( \mathcal{E}_{\text{peak}} = 0.1178 \, \text{V} \). For part (b), the peak EMF is \( \mathcal{E}_{\text{peak}} = 15066 \, \text{V} \).

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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 a changing magnetic field can induce an electromotive force (EMF) or voltage in a coil. This concept is foundational to the operation of electric generators and transformers. By rotating a coil of wire within a magnetic field, we can create an electric current. The law is mathematically expressed as:\[ \mathcal{E}_{induced} = -N \frac{d\Phi}{dt} \]where:
  • \( \mathcal{E}_{induced} \) is the induced EMF
  • \( N \) is the number of turns in the coil
  • \( \Phi \) is the magnetic flux passing through the coil
  • \( \frac{d\Phi}{dt} \) represents the rate of change of magnetic flux
This negative sign indicates Lenz鈥檚 Law, which shows the direction of the induced current will oppose the change in flux. Faraday鈥檚 Law helps us understand how to generate electricity efficiently through the motion of coils in magnetic fields.
Peak EMF Calculation
Calculating the peak electromotive force involves using a specific formula derived from Faraday鈥檚 Law. In the context of a rotating coil, the formula becomes:\[ \mathcal{E}_{peak} = N \, A \, B \, \omega \]Breaking down each component:
  • \( N \) is the number of turns in the coil.
  • \( A \) is the area of the coil, which for a square coil is calculated as side \( \times \) side.
  • \( B \) is the magnetic field strength measured in teslas (T).
  • \( \omega \), the angular velocity, is calculated using the rotational frequency: \( \omega = 2\pi f \).
Combining these factors, one can determine the peak EMF produced by the coil as it rotates through the magnetic field. This formula captures how various physical properties like the coil size and field strength directly influence the amount of electricity generated.
Magnetic Field Influence
The strength of the magnetic field, denoted as \( B \), plays a crucial role in determining the magnitude of the EMF produced by the coil. A stronger magnetic field means a greater number of magnetic field lines intersect the coil, thereby inducing a larger EMF.Consider different scenarios:
  • In Earth's magnetic field, with \( B = 5 \times 10^{-5} \, \text{T} \), the generated EMF is relatively small.
  • However, in a strong field like \( 6.4 \, \text{T} \), the EMF increases dramatically.
This demonstrates the linear relationship between the magnetic field strength and the induced EMF. As the magnetic field strength increases, so does the potential for generating electricity, showcasing why strong magnets are crucial in high-power generators.
Angular Velocity
Angular velocity, symbolized by \( \omega \), is pivotal in the generation of EMF in a rotating coil. It represents how quickly the coil turns within the magnetic field.The formula to calculate angular velocity is:\[ \omega = 2\pi f \]Here:
  • \( f \) is the frequency of rotation in hertz (Hz).
When the coil turns faster, represented by a higher frequency \( f \), the rate of change of magnetic flux through the coil increases. This results in a higher induced EMF according to Faraday's Law. In our exercise, the generator rotates at \( 100 \, \text{Hz} \), indicating rapid rotation. This high frequency amplifies the EMF generated, crucial for effective electricity production in generators.

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Most popular questions from this chapter

An \(R L C\) circuit has power factor \(0.90 .\) From this information, can you tell whether the current leads or lags the emf?

For which of the following AC circuits does impedance depend on frequency? A circuit with (a) only a resistor, (b) a resistor and inductor, (c) a resistor and capacitor.

\(\mathbf{I}=\) When a \(750-\Omega\) resistor is connected across an AC power supply, the resistor dissipates energy at a rate of \(12.5 \mathrm{~W}\). (a) Find the rms current and maximum current through the resistor. (b) What is the rms value of the power supply's emf?

A \(120-V\) (rms), 60 -Hz power supply is connected to an \(R L C\) series circuit with \(R=150 \Omega, L=1.20 \mathrm{mH},\) and \(C=33.5 \mu \mathrm{F}\) Find (a) the reactance of the capacitor and inductor, (b) the circuit impedance, and (c) the peak current.

A home receives \(1.5 \mathrm{~kW}\) of electric power from a transformer on a nearby pole. The line from the substation to the transformer is at \(10 \mathrm{kV}\) and has \(0.13 \Omega\) resistance. What fraction of the power is lost in this transmission line?
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