Ch 29: The Magnetic Field

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 29: The Magnetic Field

Problem 32

Chapter 29, Problem 32

The Hall voltage across a conductor in a 55 mT magnetic field is 1.9 μV. When used with the same current in a different magnetic field, the voltage across the conductor is 2.8 μV. What is the strength of the second field?

Verified step by step guidance

Step 1: Recall the formula for Hall voltage, which is given by \( V_H = \frac{IB}{ne} \), where \( V_H \) is the Hall voltage, \( I \) is the current, \( B \) is the magnetic field strength, \( n \) is the charge carrier density, and \( e \) is the elementary charge. In this problem, \( n \), \( e \), and \( I \) are constant for the conductor.

Step 2: Since \( n \), \( e \), and \( I \) are constant, the Hall voltage is directly proportional to the magnetic field strength. This means \( \frac{V_{H1}}{B_1} = \frac{V_{H2}}{B_2} \), where \( V_{H1} \) and \( B_1 \) are the Hall voltage and magnetic field strength for the first case, and \( V_{H2} \) and \( B_2 \) are the Hall voltage and magnetic field strength for the second case.

Step 3: Substitute the known values into the proportionality equation. For the first case, \( V_{H1} = 1.9 \mu V \) and \( B_1 = 55 \text{ mT} \). For the second case, \( V_{H2} = 2.8 \mu V \). The equation becomes \( \frac{1.9 \times 10^{-6}}{55 \times 10^{-3}} = \frac{2.8 \times 10^{-6}}{B_2} \).

Step 4: Rearrange the equation to solve for \( B_2 \). Multiply both sides by \( B_2 \) and then divide by \( \frac{1.9 \times 10^{-6}}{55 \times 10^{-3}} \) to isolate \( B_2 \). The equation becomes \( B_2 = \frac{2.8 \times 10^{-6} \cdot 55 \times 10^{-3}}{1.9 \times 10^{-6}} \).

Step 5: Perform the calculation to find \( B_2 \). Ensure that units are consistent throughout the calculation, and express the result in millitesla (mT).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Hall Effect

The Hall Effect is the phenomenon where a voltage difference (Hall voltage) is generated across a conductor when it is placed in a magnetic field perpendicular to the direction of current flow. This effect occurs due to the Lorentz force acting on the charge carriers, causing them to accumulate on one side of the conductor, thus creating a measurable voltage.

Magnetic Field Strength

Magnetic field strength, often denoted as B, is a measure of the magnetic force experienced by a charged particle in a magnetic field. It is typically measured in teslas (T) or milliteslas (mT). The strength of the magnetic field directly influences the Hall voltage produced in a conductor, as a stronger field results in a greater separation of charge carriers and thus a higher Hall voltage.

Proportionality in Hall Voltage

The Hall voltage is directly proportional to both the magnetic field strength and the current flowing through the conductor. This relationship can be expressed mathematically as V_H = k * B * I, where V_H is the Hall voltage, B is the magnetic field strength, I is the current, and k is a constant that depends on the material properties. Understanding this proportionality is essential for calculating unknown magnetic field strengths based on measured Hall voltages.

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