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Magazine Chapter 5: Problem 3
The rms voltage output of a generator is \(220 \mathrm{V}\). The coil is a square of side \(20.0 \mathrm{cm},\) has 300 turns of wire and rotates at 50 revolutions per second. What is the magnetic field?
Short Answer
Expert verified
The magnetic field strength is 0.0825 T.
Step by step solution
01
Identify known values
List all the known values from the problem:\(V_{\text{rms}} = 220 \text{ V}\), side of the square coil \(a = 20.0 \text{ cm} = 0.2 \text{ m}\), number of turns \(N = 300\), and frequency \(f = 50 \text{ Hz}\).
02
Write down the rms voltage formula
The rms voltage across a coil in a magnetic field is given by: \[ V_{\text{rms}} = N \times A \times B \times \frac{\text{d} \theta}{\text{d}t} \times \frac{1}{\text{sqrt}(2)} \]Where \(A\) is the area of the coil, \(B\) is the magnetic field strength and \(\frac{\text{d} \theta}{\text{d}t} = 2\text{{蟺}}f\).
03
Calculate the area of the coil
The area of the square coil is given by:\[A = a^2 = (0.2 \text{ m})^2 = 0.04 \text{ m}^2\]
04
Solve for the magnetic field strength
Rearrange the rms voltage formula to solve for the magnetic field strength \(B\):\[ B = \frac{V_{\text{rms}} \times \text{sqrt}(2)}{N \times A \times 2\text{蟺}f} \]Substitute the known values:\[ B = \frac{220 \text{ V} \times \text{sqrt}(2)}{300 \times 0.04 \text{ m}^2 \times 2\text{蟺} \times 50 \text{ Hz}} \]Calculate the numerator and the denominator separately before dividing them to find the result.
05
Compute the magnetic field strength
First, compute the numerator:\[ 220 \times \text{sqrt}(2) = 220 \times 1.414 = 311 \text{ V} \]Next, compute the denominator:\[ 300 \times 0.04 \times 2\text{蟺} \times 50 = 300 \times 0.04 \times 6.283 \times 50 = 3768 \text{ m}^2\text{Hz} \]Finally, divide the numerator by the denominator:\[ B = \frac{311}{3768} \text{ T} = 0.0825 \text{ T} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
rms voltage
RMS voltage, or Root Mean Square voltage, is a way of expressing AC voltage comparable to DC voltage, giving a measure of the equivalent power delivered. For an electrical generator, it's calculated to represent the effective voltage.
For a sine wave, this can be found using the formula:
\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\text{sqrt}(2)} \] Here, since we're already given the RMS voltage value, calculation is straightforward. Studying RMS voltage helps us understand how much voltage we're actually working with in practical terms.
For a sine wave, this can be found using the formula:
\[ V_{\text{rms}} = \frac{V_{\text{peak}}}{\text{sqrt}(2)} \] Here, since we're already given the RMS voltage value, calculation is straightforward. Studying RMS voltage helps us understand how much voltage we're actually working with in practical terms.
magnetic field strength
Magnetic field strength, denoted as B, is key to determining the effectiveness of a generator's coil to induce voltage. It quantifies the strength of magnetic forces in a given area. In the context of our generator problem, we use the following modified equation:
\[ B = \frac{V_{\text{rms}} \times \text{sqrt}(2)}{N \times A \times 2\text{蟺}f} \] Later, we substitute the known values to find B. This ensures we understand the interplay between variables like voltage, coil area, and frequency.
\[ B = \frac{V_{\text{rms}} \times \text{sqrt}(2)}{N \times A \times 2\text{蟺}f} \] Later, we substitute the known values to find B. This ensures we understand the interplay between variables like voltage, coil area, and frequency.
coil area
The area of the coil, denoted as A, is essential in calculating the magnetic field strength. For a square coil, the area is found using the simple formula:
\[ A = a^2 \] where 'a' is the side length. Given the side is 20.0 cm (or 0.2 m), the area becomes:
\[ A = (0.2 \text{ m})^2 = 0.04 \text{ m}^2 \] The larger the coil area, the more voltage can be induced, highlighting the importance of maximizing effective coil dimensions in generator design.
\[ A = a^2 \] where 'a' is the side length. Given the side is 20.0 cm (or 0.2 m), the area becomes:
\[ A = (0.2 \text{ m})^2 = 0.04 \text{ m}^2 \] The larger the coil area, the more voltage can be induced, highlighting the importance of maximizing effective coil dimensions in generator design.
generator frequency
Generator frequency, denoted as f, is the rate at which the coil rotates. It is measured in Hertz (Hz) and directly affects the voltage generated. In our problem, the frequency is 50 Hz. This ties into the formula as twice the product of pi and the frequency:
\[ \frac{\text{d} \theta}{\text{d}t} = 2 \text{蟺} f \] Frequency represents how many cycles occur per second and is crucial for understanding the generator's efficiency. Higher frequency generally means higher induced voltage.
\[ \frac{\text{d} \theta}{\text{d}t} = 2 \text{蟺} f \] Frequency represents how many cycles occur per second and is crucial for understanding the generator's efficiency. Higher frequency generally means higher induced voltage.
number of turns in wire
The number of turns in the wire, denoted as N, represents how many loops the wire forms in the coil. Each turn increases the induced voltage as it multiplies the interaction with the magnetic field. In our case, there are 300 turns. High numbers of turns enhance the magnetic interaction and the eventual voltage output:
\[ V_{\text{rms}} = N \times A \times B \times \frac{\text{d} \theta}{\text{d}t} \times \frac{1}{\text{sqrt}(2)} \] This emphasizes that more turns improve the coil's capability to convert magnetic energy into an electrical signal, making them a critical design parameter.
\[ V_{\text{rms}} = N \times A \times B \times \frac{\text{d} \theta}{\text{d}t} \times \frac{1}{\text{sqrt}(2)} \] This emphasizes that more turns improve the coil's capability to convert magnetic energy into an electrical signal, making them a critical design parameter.
