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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 58
Chapter 29, Problem 58

A proton moving in a uniform magnetic field with v⃗1=(1.00×106i^)m/s\(\vec{v}\)_1 = (1.00 \(\times\) 10^6\,\(\hat{i}\))\,\(\text{m/s}\) experiences force F⃗1=(1.20×10−16k^)N\(\vec{F}\)_1 = (1.20 \(\times\) 10^{-16}\,\(\hat{k}\))\,\(\text{N}\). A second proton with v⃗2=(2.0×106j^)m/s\(\vec{v}\)_2 = (2.0 \(\times\) 10^6\,\(\hat{j}\))\,\(\text{m/s}\) experiences F⃗2=(−4.16×10−16k^)N\(\vec{F}\)_2 = (-4.16 \(\times\) 10^{-16}\,\(\hat{k}\))\,\(\text{N}\) in the same field. What is B⃗\(\vec{B}\)? Give your answer as a magnitude and an angle measured counter-clockwise from the +x+x-axis.

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Step 1: Recall the formula for the magnetic force on a charged particle moving in a magnetic field: \( \vec{F} = q \vec{v} \times \vec{B} \), where \( \vec{F} \) is the magnetic force, \( q \) is the charge of the particle, \( \vec{v} \) is the velocity vector, and \( \vec{B} \) is the magnetic field vector.
Step 2: Use the cross product formula \( \vec{v} \times \vec{B} \) to relate the given velocity vectors \( \vec{v}_1 \) and \( \vec{v}_2 \) to their respective force vectors \( \vec{F}_1 \) and \( \vec{F}_2 \). For \( \vec{v}_1 = (1.00 \times 10^6 \hat{i}) \text{ m/s} \) and \( \vec{F}_1 = (1.20 \times 10^{-16} \hat{k}) \text{ N} \), the cross product simplifies to \( \vec{F}_1 = q \vec{v}_1 \times \vec{B} \). Similarly, for \( \vec{v}_2 = (2.0 \times 10^6 \hat{j}) \text{ m/s} \) and \( \vec{F}_2 = (-4.16 \times 10^{-16} \hat{k}) \text{ N} \), use \( \vec{F}_2 = q \vec{v}_2 \times \vec{B} \).
Step 3: Solve for \( \vec{B} \) using the known charge of a proton \( q = 1.60 \times 10^{-19} \text{ C} \). For \( \vec{F}_1 \), substitute \( \vec{v}_1 \) and \( \vec{F}_1 \) into the cross product equation to find the components of \( \vec{B} \). Similarly, use \( \vec{v}_2 \) and \( \vec{F}_2 \) to find the components of \( \vec{B} \).
Step 4: Combine the results from both cases to determine the magnetic field vector \( \vec{B} \). The components of \( \vec{B} \) can be expressed as \( B_x \) and \( B_y \), which are derived from the relationships between \( \vec{v}_1 \), \( \vec{v}_2 \), \( \vec{F}_1 \), and \( \vec{F}_2 \).
Step 5: Calculate the magnitude of \( \vec{B} \) using \( |\vec{B}| = \sqrt{B_x^2 + B_y^2} \), and determine the angle \( \theta \) of \( \vec{B} \) relative to the +x-axis using \( \theta = \tan^{-1}(B_y / B_x) \). Express the angle in counter-clockwise direction from the +x-axis.

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Key Concepts

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

Lorentz Force

The Lorentz force describes the force experienced by a charged particle moving through a magnetic field. It is given by the equation \\vec{F} = q(\\vec{v} \times \\vec{B})\\, where \\vec{F} is the force, \\vec{v} is the velocity of the particle, \\vec{B} is the magnetic field, and q is the charge of the particle. The direction of the force is determined by the right-hand rule, which states that if you point your fingers in the direction of \\vec{v} and curl them towards \\vec{B}, your thumb will point in the direction of \\vec{F}.
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Magnetic Field

A magnetic field is a vector field that exerts a force on moving charges and magnetic dipoles. It is represented by the symbol \\vec{B} and is measured in teslas (T). The magnetic field can be uniform or vary in space, and its direction is defined as the direction a north pole of a magnet would point. Understanding the properties of the magnetic field is crucial for analyzing the motion of charged particles within it.
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Vector Components

Vectors can be broken down into their components along the coordinate axes, typically x, y, and z. For example, a vector \\vec{v} can be expressed as \\vec{v} = v_x \\hat{i} + v_y \\hat{j} + v_z \\hat{k}\\, where \\hat{i}, \\hat{j}, and \\hat{k} are the unit vectors in the x, y, and z directions, respectively. This decomposition is essential for solving problems in physics, as it allows for the application of vector addition and the analysis of forces and motions in multiple dimensions.
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Related Practice
Textbook Question

An antiproton (same properties as a proton except that q = -e) is moving in the combined electric and magnetic fields of FIGURE P29.61. What are the magnitude and direction of the antiproton's acceleration at this instant?

Textbook Question

A long, hollow wire has inner radius R₁ and outer radius R₂. The wire carries current I uniformly distributed across the area of the wire. Use Ampère's law to find an expression for the magnetic field strength in the three regions 0 < r < R₁, R₁ < r < R₂, and R₂ < r.

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Textbook Question

The toroid of FIGURE P29.54 is a coil of wire wrapped around a doughnut-shaped ring (a torus). Toroidal magnetic fields are used to confine fusion plasmas. Is a toroidal magnetic field a uniform field? Explain.

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Textbook Question

A 65-cm-diameter cyclotron uses a 500 V oscillating potential difference between the dees. What is the maximum kinetic energy of a proton if the magnetic field strength is 0.75 T?

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Textbook Question

A flat, circular disk of radius R is uniformly charged with total charge Q. The disk spins at angular velocity ω about an axis through its center. What is the magnetic field strength at the center of the disk?

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Textbook Question

An electron in a cathode-ray tube is accelerated through a potential difference of 10 kV, then passes through the 2.0-cm-wide region of uniform magnetic field in FIGURE P29.60. What field strength will deflect the electron by 10°?