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Chapter 11: Problem 25

Integrated Concepts To construct a nonmechanical 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 in liters per second through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV. (b) What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?

Short Answer

Expert verified
The flow rate through a 3.00-cm-diameter pipe is 2.82 liters per second. The Hall voltage for the same flow rate through a 10.0-cm-diameter pipe is 17.9 mV.

Step by step solution

01

Understanding the Hall Effect for Fluid Flow Measurement

The Hall effect can be used to measure fluid flow by applying a magnetic field perpendicular to the flow direction and measuring the induced voltage (Hall voltage) across the fluid. Given a magnetic field (B), we can find the flow velocity (v) using the relation for Hall voltage (V_H), which is given by the equation V_H = Bvd, where d is the diameter of the pipe and v is the flow velocity.
02

Calculating the Flow Velocity

To calculate the flow velocity, rearrange the Hall voltage equation: v = V_H / (Bd). Substitute the given values: B = 0.500 T, V_H = 60.0 mV = 0.060 V, and d = 3.00 cm = 0.030 m. The flow velocity (v) then becomes: v = 0.060 V / (0.500 T * 0.030 m) = 4.0 m/s.
03

Finding the Flow Rate in Liters per Second

Flow rate (Q) in cubic meters per second can be found by multiplying the flow velocity by the cross-sectional area of the pipe (A = \( \pi r^2 \)): Q = Av. Since the diameter (d) is 3.00 cm, the radius (r) is 1.50 cm or 0.015 m. Thus, Q = \( \pi (0.015 m)^2 \) * 4.0 m/s = 2.82 \( \times 10^{-3} \) m^3/s. To convert to liters per second, use the conversion 1 m^3/s = 1000 L/s, resulting in a flow rate of 2.82 L/s.
04

Using the Flow Rate to Find the Hall Voltage for a Different Pipe Diameter

For a 10.0 cm-diameter pipe, we must first find the flow velocity using the original flow rate. Since Q remains constant, Q = Av, and for the new diameter (10.0 cm or 0.1 m), the radius changes to 0.05 m. We rearrange to v = Q / A = Q / (\( \pi r^2 \)). Substituting in the values, v = 2.82 \( \times 10^{-3} \) m^3/s / \( \pi (0.05 m)^2 \) = 0.359 m/s. To find the new Hall voltage (V_H'), use the relation V_H' = Bvd. With the flow velocity and the diameter for the new pipe, V_H' = 0.500 T * 0.359 m/s * 0.1 m = 17.9 mV.

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Key Concepts

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

Magnetic Field in Hall Effect Flow Measurement
Understanding the role of a magnetic field is critical in the Hall effect water flow measurement technique. A magnetic field, which we measure in teslas (T), is applied perpendicular to the flow of the water in a pipe. It's this field that interact with the moving charges in the water, leading to the generation of the Hall voltage across the flow direction. This field has to be uniform and stable, so our flow measurements are accurate. In our exercise, a magnetic field of 0.500 tesla is strong enough to cause the necessary interaction without affecting the water properties. While studying magnetic fields, remember that the stronger the field, the larger the induced Hall voltage for a given flow rate, provided other conditions remain constant.
Hall Voltage
The Hall voltage is the electrical potential difference generated by the magnetic field's influence on the charges in the flowing water. It's the heart of this measurement technique and what we use to determine the flow rate of water. When water or any conductive fluid passes through a magnetic field, charged particles are deflected to one side, creating this voltage.In our exercise, with a fixed magnetic field and a known pipe diameter, the generated Hall voltage gives direct information about the flow velocity of the water. We're given a Hall voltage of 60.0 millivolts (mV) initially. This voltage is directly proportional to both the strength of the magnetic field and the velocity of the moving water, making it an extremely useful parameter in calculating flow rate.
Flow Rate Calculation
Flow rate calculation is vital to understand how much water is moving through the pipe. It's typically measured in cubic meters per second (m鲁/s) or liters per second (L/s), with the relation between them being 1 m鲁/s is equivalent to 1000 L/s.To calculate flow rate from Hall voltage, you use the formula
\( Q = Av \)where \( Q \) is the flow rate, \( A \) is the cross-sectional area, and \( v \) is the flow velocity obtained from the Hall voltage equation \( v = \frac{V_H}{Bd} \). The conversion from cubic meters to liters makes the results much more relatable, as liters are a more common volume measurement in everyday life.
Cross-sectional Area in Flow Measurement
The cross-sectional area of the pipe is a determining factor in both the induced Hall voltage and the calculation of the flow rate. It's the area through which the water travels, and for a circular pipe, it can be found using the formula for the area of a circle, \( A = \pi r^2 \), where \( r \) is the pipe's radius.In flow calculations, the cross-sectional area must be squared, dramatically affecting the results for the flow rate. A small change in diameter leads to a significant change in the cross-sectional area, and consequently, in the velocity and Hall voltage required to maintain the same flow rate. For instance, when the problem transitions from a smaller diameter pipe to a larger one, there's a substantial drop in the required Hall voltage to maintain the same flow rate, due to the increase in cross-sectional area.

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Most popular questions from this chapter

Find the current through a loop needed to create a maximum torque of \(9.00 \mathrm{~N} \cdot \mathrm{m}\). The loop has 50 square turns that are \(15.0 \mathrm{~cm}\) on a side and is in a uniform \(0.800-\mathrm{T}\) magnetic field.

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