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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
All textbooksKnight Calc 5th EditionCh 29: The Magnetic FieldProblem 78
Chapter 29, Problem 78

A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a 6.5-m-long, 2.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.0 mT magnetic field 1.0 mm from the surface of the wire. What is the material's resistivity?

Verified step by step guidance
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Step 1: Understand the problem. The goal is to calculate the resistivity (ρ) of the material. Resistivity is related to the resistance (R) of the wire by the formula: ρ=RA/L, where A is the cross-sectional area of the wire and L is its length. To find R, we will use Ohm's law: V=IR, where V is the voltage and I is the current. The current I can be determined using the magnetic field created by the wire.
Step 2: Use Ampere's law to find the current (I). The magnetic field (B) around a long, straight wire carrying current I is given by: B=μ0I/(2πr), where μ0 is the permeability of free space (4π×10-7 T·m/A), and r is the distance from the center of the wire to the point where the magnetic field is measured. Rearrange this formula to solve for I: I=2πrB/μ0.
Step 3: Calculate the cross-sectional area (A) of the wire. The wire is cylindrical, so its cross-sectional area is given by: A=π(d/2)2, where d is the diameter of the wire. Substitute the given diameter (2.0 mm) into this formula to find A.
Step 4: Use Ohm's law to find the resistance (R) of the wire. Rearrange the formula V=IR to solve for R: R=V/I. Substitute the given voltage (1.5 V) and the current (I) calculated in Step 2 into this formula.
Step 5: Calculate the resistivity (ρ) of the material. Use the formula ρ=RA/L, where R is the resistance from Step 4, A is the cross-sectional area from Step 3, and L is the length of the wire (6.5 m). Substitute these values into the formula to find the resistivity.

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Key Concepts

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

Resistivity

Resistivity is a fundamental property of materials that quantifies how strongly they resist the flow of electric current. It is denoted by the symbol ρ (rho) and is measured in ohm-meters (Ω·m). The resistivity of a material depends on its composition and temperature, and it is crucial for determining how much voltage is needed to drive a current through a given length and cross-sectional area of the material.
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Magnetic Field Around a Current-Carrying Wire

When an electric current flows through a wire, it generates a magnetic field around it. The strength of this magnetic field can be calculated using Ampère's Law, which relates the magnetic field to the current and the distance from the wire. In this scenario, the scientist measures the magnetic field at a specific distance from the wire, which can help infer the current flowing through the wire and, subsequently, the resistivity of the material.
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Ohm's Law

Ohm's Law is a fundamental principle in electrical engineering and physics that states the relationship between voltage (V), current (I), and resistance (R) in a circuit. It is expressed as V = I × R. This law is essential for calculating the current flowing through the wire when a voltage is applied, which is necessary for determining the resistivity of the material based on the wire's dimensions and the measured magnetic field.
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Related Practice
Textbook Question

A long, straight conducting wire of radius R has a nonuniform current density J = J₀r/R, where J₀ is a constant. The wire carries total current I. Find an expression for the magnetic field strength inside the wire at radius r.

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Textbook Question

A wire along the x-axis carries current I in the negative x-direction through the magnetic field B⃗={B0xlk^0≤x≤l0elsewhere\(\vec{B}\)= \(\begin{cases}\) B_0\(\dfrac{x}{l}\]\hat{k}\) & 0 \(\leq\) x \(\leq\) l \\ 0 & \(\text{elsewhere}\) \(\end{cases}\). Find an expression for the net torque on the wire about the point x = 0.

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Textbook Question

FIGURE CP29.79 is an edge view of a 2.0 kg square loop, 2.5 m on each side, with its lower edge resting on a frictionless, horizontal surface. A 25 A current is flowing around the loop in the direction shown. What is the strength of a uniform, horizontal magnetic field for which the loop is in static equilibrium at the angle shown?

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Textbook Question

Determine the field strength at the center of a current-carrying square loop having sides of length 2R.

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Textbook Question

An electromagnetic rail gun uses magnetic forces to launch projectiles. FIGURE P29.76 shows a 10-cm-long, 10 g metal wire that can slide without friction along 1.0-m-long horizontal rails. The rails are connected to a 300 V source, and a 0.10 T magnetic field fills the space between the rails. Each rail has linear resistivity ⋋ = 0.10 Ω/m, which means that the resistance is ⋋ multiplied by the length of rail through which current flows. Assume that the sliding wire and the left end, where the voltage source is, have zero resistance. The wire is initially placed at x₀ = 5.0 cm then the switch is closed. What is the wire's speed as it leaves the rails?

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Textbook Question

In FIGURE P29.75, a long, straight, current-carrying wire of linear mass density μ is suspended by threads. A magnetic field perpendicular to the wire exerts a horizontal force that deflects the wire to an equilibrium angle θ. Find an expression for the strength and direction of the magnetic field B.

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