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(II) A 250-loop circular armature coil with a diameter of \(10.0 \mathrm{~cm}\) rotates at \(120 \mathrm{rev} / \mathrm{s}\) in a uniform magnetic field of strength \(0.45 \mathrm{~T}\). What is the rms voltage output of the generator? What would you do to the rotation frequency in order to double the rms voltage output?

Short Answer

Expert verified
The rms voltage output is 472.73 volts. Double the rotation frequency to 240 rev/s to double the rms voltage.

Step by step solution

01

Calculate the area of the coil

The area (A) of the coil is calculated using the formula for the area of a circle \( A = \pi r^2 \), where \( r \) is the radius of the coil. Given the diameter of the coil is 10.0 cm, the radius \( r = \frac{10.0}{2} = 5.0 \) cm = 0.05 m.So, \( A = \pi (0.05)^2 = 0.00785 \) m\(^2\).
02

Determine the angular speed of the coil

The angular speed \( \omega \) is calculated from the rotation frequency using the formula \( \omega = 2\pi f \), where \( f = 120 \) rev/s is the frequency.Thus, \( \omega = 2\pi \times 120 = 240\pi \) rad/s.
03

Calculate the peak voltage of the generator

The peak voltage (\( V_0 \)) is given by the formula \( V_0 = NAB\omega \), where \( N = 250 \) is the number of loops, \( A = 0.00785 \) m\(^2\) is the area, \( B = 0.45 \) T is the magnetic field strength, and \( \omega = 240\pi \) rad/s.So, \( V_0 = 250 \times 0.00785 \times 0.45 \times 240\pi = 668.77 \) volts.
04

Calculate the rms voltage output

The rms voltage (\( V_{rms} \)) is calculated as \( V_{rms} = \frac{V_0}{\sqrt{2}} \).Thus, \( V_{rms} = \frac{668.77}{\sqrt{2}} = 472.73 \) volts.
05

Determine the required frequency to double the rms voltage output

To double the rms voltage output, we need to double the peak voltage \( V_0 \), which can be done by doubling the angular speed \( \omega \). Since \( \omega = 2\pi f \), doubling \( \omega \) implies doubling the frequency \( f \).Therefore, the new frequency should be \( 2 \times 120 = 240 \) rev/s.

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Key Concepts

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

RMS Voltage
Root mean square (RMS) voltage is a crucial concept in understanding how voltage behaves in alternating current (AC) circuits. Essentially, it provides an average value of alternating voltage, making it comparable to direct current (DC) voltage in terms of energy delivery. RMS voltage is particularly useful when dealing with devices that operate on AC because it helps predict the amount of power a circuit can deliver.

**Why It’s Important**
The peak voltage (\( V_0 \)) is not constant over time in an AC system. Instead, it varies, reaching its maximum in both positive and negative directions. The RMS voltage simplifies this oscillation into a single, constant value which makes calculations and comparisons easier.
  • The RMS value of voltage ensures that the same heating effect is produced in a resistor, as a DC voltage of the same value would do.
  • It simplifies complex AC waveforms into a single value that reflects the effective voltage.
In the given problem, once the peak voltage is calculated (\( V_0 = 668.77 \) volts), the RMS voltage is simply \( V_{rms} = \frac{V_0}{\sqrt{2}} = 472.73 \) volts.

Understanding RMS voltage helps in efficiently analyzing electrical systems and ensures correct operation of electrical devices and systems that utilize AC voltage.
Angular Speed
Angular speed is a fundamental concept when studying rotating systems, especially in the context of electromagnetic induction. It describes how quickly an object is rotating. Specifically, it tells us the angle through which an object turns in a given time period, typically measured in radians per second.

**Connection to Frequency**
In the context of a generator, angular speed (\( \omega \)) is directly related to how fast the coil rotates. It's calculated using the relationship \( \omega = 2\pi f \), where \( f \) is the frequency in revolutions per second. Therefore:
  • When the frequency is given as \( f = 120 \) rev/s, \( \omega = 2\pi \times 120 = 240\pi \) rad/s.
This angular speed is essential because it directly affects the electromotive force (EMF) and subsequent voltage produced in the generator.

**Significance in Voltage Generation**
A higher angular speed results in a greater rate of change of magnetic flux, leading to increased voltage output. Therefore, to double the RMS voltage, as posed in the problem, you would need to double the frequency, which in turn doubles angular speed. This shows how closely angular speed ties to voltage production efficiency.
Magnetic Field Strength
The concept of magnetic field strength is crucial when diving into electromagnetic induction. It measures the intensity of a magnetic field, usually highlighted in teslas (T). A stronger magnetic field creates a more significant effect on moving charges and, consequently, on current-carrying conductors in its presence.

**Impact on Voltage Generation**
Magnetic field strength (\( B \)) plays a core role in determining the voltage output of generators. Its relationship with voltage is evident in the formula used to calculate peak voltage:
  • \( V_0 = NAB\omega \), where \( N \) is the number of loops, \( A \) is the area of the coil, \( B \) is the magnetic field strength, and \( \omega \) is the angular speed.
In the given exercise, a magnetic field strength of \( 0.45 \) T is provided, contributing significantly to the calculated peak voltage.

**Why It Matters**
The stronger the magnetic field, the greater the induced EMF and thus the higher the voltage output. In practical applications, the magnetic field strength can be adjusted or selected according to the desired voltage output, thus making it a vital factor in designing electromagnetic systems.
This highlights its pivotal role in electromagnetic induction processes and technologies. Understanding how \( B \) affects power output can guide optimizing and enhancing systems using magnetic fields.

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Most popular questions from this chapter

(I) A 22.0-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in \(0.20 \mathrm{~s}\) s. What is the average induced emf in the loop?

The primary windings of a transformer which has an \(85 \%\) efficiency are connected to 110-V ac. The secondary windings are connected across a \(2.4-\Omega, 75-\mathrm{W}\) lightbulb. (a) Calculate the current through the primary windings of the transformer. (b) Calculate the ratio of the number of primary windings of the transformer to the number of secondary windings of the transformer.

(II) We are looking down on an elastic conducting loop with resistance \(R=2.0 \Omega,\) immersed in a magnetic field. The field's magnitude is uniform spatially, but varies with time \(t\) according to \(B(t)=\alpha t,\) where \(\alpha=0.60 \mathrm{~T} / \mathrm{s} .\) The area \(A\) of the loop also increases at a constant rate, according to \(A(t)=A_{0}+\beta t,\) where \(A_{0}=0.50 \mathrm{~m}^{2}\) and \(\beta=0.70 \mathrm{~m}^{2} / \mathrm{s}\) Find the magnitude and direction (clockwise or counterclockwise, when viewed from above the page) of the induced current within the loop at time \(t=2.0 \mathrm{~s}\) if the magnetic field \((a)\) is parallel to the plane of the loop to the right; \((b)\) is perpendicular to the plane of the loop, down.

(II) A single circular loop of wire is placed inside a long solenoid with its plane perpendicular to the axis of the solenoid. The area of the loop is \(A_{1}\) and that of the solenoid, which has \(n\) turns per unit length, is \(A_{2} .\) A current \(I=I_{0} \cos \omega t\) flows in the solenoid turns. What is the induced emf in the small loop?

A circular metal disk of radius \(R\) rotates with angular velocity \(\omega\) about an axis through its center perpendicular to its face. The disk rotates in a uniform magnetic field \(B\) whose direction is parallel to the rotation axis. Determine the emf induced between the center and the edges.

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