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Chapter 11: Problem 98

To construct a non-mechanical water meter, a 0.500-T magnetic field is placed across the supply water pipe to a home and the Hall voltage is recorded. (a) Find the flow rate through a 3.00 - cm-diameter pipe if the Hall voltage is \(60.0 \mathrm{mV}\). (b) What would the Hall voltage be for the same flow rate through a \(10.0-\mathrm{cm}-\) diameter pipe with the same field applied?

Short Answer

Expert verified
The flow rate through a 3.00-cm-diameter pipe is approximately \(2.83 \times 10^{-3} \mathrm{m^3/s}\). For the same flow rate through a 10.0-cm-diameter pipe with the same magnetic field applied, the Hall voltage would be approximately 0.018 V (18.0 mV).

Step by step solution

01

Identify given values

In this problem, we are given: - Magnetic field (B): 0.500 T - Hall voltage (V_H): 60.0 mV (0.060 V) - Pipe diameter (D): 3.00 cm (0.03 m)
02

Calculate linear velocity

Use the Hall Voltage relationship to calculate the linear velocity (v): \(V_H = B \cdot v \cdot d\) Rearrange for linear velocity (v): \(v = \frac{V_H}{B \cdot d}\) Plug in the given values: \(v = \frac{0.060}{0.500 \cdot 0.03} = 4 \mathrm{m/s}\)
03

Calculate cross-sectional area

Use the formula for cross-sectional area (A) of a circular pipe: \(A = \pi \cdot (D / 2)^2\) Plug in the given diameter (D = 0.03 m): \(A = \pi \cdot (0.03 / 2)^2 鈮� 7.07 \times 10^{-4} \mathrm{m^2}\)
04

Calculate flow rate

Use the flow rate (Q) and linear velocity (v) relationship: \(Q = A \cdot v\) Plug in the calculated linear velocity (v) and cross-sectional area (A): \(Q = 7.07 \times 10^{-4} \cdot 4 鈮� 2.83 \times 10^{-3} \mathrm{m^3/s}\) So, the flow rate through a 3.00-cm-diameter pipe is approximately \(2.83 \times 10^{-3} \mathrm{m^3/s}\). #b) What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?#
05

Calculate cross-sectional area for the larger pipe

Using the given diameter (D) for the larger pipe (10.0 cm or 0.1 m), calculate the new cross-sectional area (A'): \(A' = \pi \cdot (0.1 / 2)^2 鈮� 7.85 \times 10^{-3} \mathrm{m^2}\)
06

Calculate linear velocity for the larger pipe

Use the flow rate (Q) and cross-sectional area (A') to find the new linear velocity (v'): \(Q = A' \cdot v'\) Rearrange for linear velocity (v'): \(v' = \frac{Q}{A'}\) Plug in the previous flow rate (Q) and the new cross-sectional area (A'): \(v' = \frac{2.83 \times 10^{-3}}{7.85 \times 10^{-3}} 鈮� 0.36 \mathrm{m/s}\)
07

Calculate the new Hall voltage

Use the relationship between Hall voltage (V_H), magnetic field (B), linear velocity (v'), and pipe diameter (D) for the larger pipe: \(V_H' = B \cdot v' \cdot d'\) Plug in the given magnetic field (0.500 T) and diameter (D' = 0.1 m), along with the calculated linear velocity (v'): \(V_H' = 0.500 \cdot 0.36 \cdot 0.1 鈮� 0.018 \mathrm{V}\) So, the new Hall voltage for the same flow rate through the larger 10.0-cm-diameter pipe would be approximately 0.018 V (18.0 mV).

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Key Concepts

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

Magnetic Field
Learn about the invisible force that facilitates the working of a Hall voltage water meter. In our example, a magnetic field of 0.500 teslas (T) is essential for the operation of the device. But what exactly is a magnetic field?

A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. This magnetic field can be visualized as a pattern of invisible lines that exit a magnet's north pole and enter its south pole. These lines represent the path that magnetic forces follow. In this scenario, the strength of the magnetic field, usually measured in teslas or gauss, is critical since it is directly proportional to the Hall voltage that is generated.

The principle behind measuring flow rate using a magnetic field relies on Faraday's law of electromagnetic induction, which indicates that a voltage is induced when a conductor moves through a magnetic field. The Hall voltage is a manifestation of this physical principle, providing a non-invasive method to gauge the flow of water inside a pipe.
Flow Rate Calculation
Understanding flow rate calculation is crucial for interpreting results from a Hall voltage water meter. The flow rate indicates how much fluid moves past a point within a certain period. It is the volume of fluid flowing per unit of time, usually expressed in liters per second (L/s) or cubic meters per second (m鲁/s).

To calculate the flow rate (Q) in our exercise, two pieces of information are used: the linear velocity (v) of the water and the cross-sectional area (A) of the pipe. Mathematically, the flow rate is found by multiplying these two quantities:
\[Q = A \cdot v\]
Once the cross-sectional area is known by calculating the radius of the pipe (half of the diameter), and the linear velocity is determined from the Hall voltage and magnetic field, these values are inserted into the equation to find the flow rate. This step is vital in converting the observed Hall voltage to a quantifiable rate of water movement, essential in practical applications such as billing for water usage.
Electromagnetism
Delving into electromagnetism will reveal the foundational science behind the Hall voltage water meter. Electromagnetism is one of the four fundamental forces of nature and is the force responsible for both electric fields and magnetic fields. Through the relationship between these two fields, a myriad of technological innovations, including the operation of Hall voltage meters, have been made possible.

When an electric current passes through a conductor and is exposed to a magnetic field, it experiences a force (known as the Lorentz force), which is used in the Hall effect to create a measurable voltage. This relationship between electricity and magnetism underpins the calculation of flow rate in a Hall voltage water meter. In simplified terms, as water (a conductor when it contains impurities or ions) flows through a pipe within a magnetic field, the charges in the water are deflected to one side of the pipe, creating a difference in potential 鈥� the Hall voltage. This voltage is directly proportional to the flow rate of the water, allowing us to calculate how much water is traveling through the pipe.

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Most popular questions from this chapter

What magnetic field is required in order to confine a proton moving with a speed of \(4.0 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a circular orbit of radius \(10 \mathrm{cm} ?\)

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