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A Hall probe calibrated to read \(1.00 \mu \mathrm{V}\) when placed in a 2.00 -T field is placed in a 0.150 -T field. What is its output voltage?

Short Answer

Expert verified
The output voltage is 0.075 μV in the 0.150-T field.

Step by step solution

01

Understand the Relationship Between Voltage and Magnetic Field

According to the problem, the Hall probe's output voltage is directly proportional to the magnetic field it is placed in. It means that if we double the magnetic field strength, we double the output voltage. Based on this, we can say that the voltage output is given by the formula: V = k * B, where k is the proportional constant for the particular Hall probe, B is the magnetic field, and V is the output voltage.
02

Calculate the Proportional Constant (k)

Using the calibration information, we can find the value of the proportional constant k. We are told that when the magnetic field (B) is 2.00 T, the output voltage (V) is 1.00 μV. Plugging these values into the formula V = k * B gives us 1.00 μV = k * 2.00 T. Solving for k gives us k = (1.00 μV) / (2.00 T).
03

Solve for the Output Voltage in the 0.150-T Field

Now that we have the constant k, we can use it to find the output voltage when the probe is in the 0.150-T field. We use the formula V = k * B where B is now 0.150 T. Substitute the value of k from Step 2 to find the new voltage output.
04

Calculate the New Output Voltage

Substituting the values into the V = k * B formula, we get V = (1.00 μV / 2.00 T) * 0.150 T. Perform the multiplication to find the output voltage V in microvolts.

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Key Concepts

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

Hall probe
A Hall probe is a device designed to measure the magnitude of a magnetic field. It operates based on the Hall effect, which occurs when a current-carrying conductor is placed perpendicular to a magnetic field, resulting in a voltage (the Hall voltage) perpendicular to both the current and the field. The Hall probe consists of a thin strip of conductive material through which an electric current is passed. When this probe is exposed to a magnetic field, the charged particles in the current experience a force, leading to a build-up of charge on one side of the conductor. This separation of charge creates an electric field and hence, a measurable voltage across the conductor.

For accurate measurements, the Hall probe is usually calibrated against known magnetic field strengths, which allows users to deduce the strength of unknown magnetic fields by comparing the output voltages. The relationship between the output voltage and the magnetic field strength is typically linear, which implies that the output voltage of a well-calibrated Hall probe can be a precise indicator of magnetic field intensity.
Magnetic field
A magnetic field is a vector field that surrounds magnets and electric currents, and it's characterized by the force it exerts on moving charged particles and magnetic materials. Fields are often visualized as lines of force that emanate from magnetic poles and circulate back to them. Magnetic fields are measured in units of teslas (T) in the International System of Units (SI).

Understanding the nature of magnetic fields is crucial for various applications, including electrical engineering, where they are fundamental to the operation of transformers, motors, and generators. Additionally, magnetic fields play an essential role in medical imaging techniques, such as Magnetic Resonance Imaging (MRI). Grasping the concept of the magnetic field is significant in physics as it helps explain the behavior of charged particles and the interrelation with electromagnetism.
Proportional constant
The proportional constant (often denoted by the letter 'k' in equations) is a factor that relates two quantities that are directly proportional to each other. In the context of Hall effect voltage calculation, the proportional constant connects the magnetic field strength to the output voltage of the Hall probe. This constant is dependent on the characteristics of the Hall probe, such as the material of the probe and its geometry.

The determination of the proportional constant is a crucial step because it allows for the conversion between the Hall voltage and the magnetic field strength. Once the constant is known, the output voltage for any other magnetic field intensity can be calculated by a simple multiplication. The consistency of this constant is key to ensuring that the measurements taken by the Hall probe remain accurate and reliable over time.
Physics problem solving
Physics problem solving is a methodical approach that involves understanding the problem, identifying known and unknown variables, and finding a systematic way to use the relationships between these variables to solve for the unknowns. The steps in the solution provided for calculating Hall effect voltage exemplify this structured approach:

  1. Recognize the relationship between the observed phenomena—in this case, the direct proportionality between voltage and magnetic field strength.
  2. Determine the proportional constant using given calibration data.
  3. Use the proportional constant to solve for the unknown—in this case, the output voltage in a different magnetic field.
Mastering physics problem solving enables students to tackle a wide range of problems by applying universal principles, enabling them to connect theoretical understanding with practical application.

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Most popular questions from this chapter

A cosmic ray electron moves at \(7.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) perpendicular to the Earth's magnetic field at an altitude where field strength is \(1.00 \times 10^{-5} \mathrm{T}\). What is the radius of the circular path the electron follows?

Is the decay \(n \rightarrow e^{+}+e^{-}\) possible considering the appropriate conservation laws? State why or why not.

(a) What is the speed of a supersonic aircraft with a 17.0-m wingspan, if it experiences a 1.60 -V Hall voltage between its wing tips when in level flight over the north magnetic pole, where the Earth's field strength is \(8.00 \times 10^{-5} \mathrm{T} ?\) (b) Explain why very little current flows as a result of this Hall voltage.

Synchrotron radiation takes energy from an accelerator beam and is related to acceleration. Why would you expect the problem to be more severe for electron accelerators than proton accelerators?

(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are \(3.90 \times 10^{-25} \mathrm{kg}\) and \(3.95 \times 10^{-25} \mathrm{kg},\) respectively, and they travel at \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a 0.250 -T field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium- 235 from uranium- 238 .

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