Problem 1 - Fundamentals of Physics

Question

A proton traveling at \(23.0^{\circ}\) with respect to the direction of a magnetic field of strength \(2.60 \mathrm{mT}\) experiences a magnetic force of \(6.50 \times 10^{-17} \mathrm{~N}\). Calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.

Step-by-Step Solution

Answer

(a) \( v \approx 1.073 \times 10^7 \mathrm{~m/s} \), (b) \( KE \approx 3.22 \times 10^5 \mathrm{~eV} \).

Step 1: Understand the Magnetic Force

The magnetic force experienced by a charge moving in a magnetic field is given by the formula: \[F = qvB \sin \theta\]where:- \( F \) is the magnetic force,- \( q \) is the charge of the proton (\(1.60 \times 10^{-19} \mathrm{~C}\)),- \( v \) is the speed of the proton,- \( B \) is the magnetic field strength (\(2.60 \mathrm{~mT} = 2.60 \times 10^{-3} \mathrm{~T}\)),- \( \theta \) is the angle between the velocity and the magnetic field (\(23.0^{\circ}\)).

Step 2: Solve for Proton's Speed

Rearrange the formula to solve for the speed \( v \):\[v = \frac{F}{qB \sin \theta}\]Substitute the known values:\[v = \frac{6.50 \times 10^{-17} \mathrm{~N}}{1.60 \times 10^{-19} \mathrm{~C} \times 2.60 \times 10^{-3} \mathrm{~T} \times \sin(23.0^{\circ})}\]Calculate \( \sin(23.0^{\circ}) \) and then compute \( v \).

Step 3: Convert Proton's Speed Calculation

Calculate the value:\[\sin(23.0^{\circ}) \approx 0.3907\]Substitute back into the equation:\[v = \frac{6.50 \times 10^{-17}}{1.60 \times 10^{-19} \times 2.60 \times 10^{-3} \times 0.3907} \approx 1.073 \times 10^7 \mathrm{~m/s}\]

Step 4: Calculate Proton's Kinetic Energy

Kinetic energy \( KE \) is given by:\[KE = \frac{1}{2} mv^2\]Substituting the mass of a proton, \( m = 1.67 \times 10^{-27} \mathrm{~kg} \), and the speed:\[KE = \frac{1}{2} \times 1.67 \times 10^{-27} \times (1.073 \times 10^7)^2\]Compute the numerical value.

Step 5: Convert Kinetic Energy to Electron-volts

Convert the kinetic energy from joules to electron-volts using the conversion factor \(1 \mathrm{~eV} = 1.60 \times 10^{-19} \mathrm{~J}\):\[KE\text{ in eV} = \frac{KE \text{ in Joules}}{1.60 \times 10^{-19}}\]Calculate the numerical value of the kinetic energy in electron-volts.

Key Concepts

Proton Speed

When a proton travels through a magnetic field, its speed is influenced by the magnetic force acting on it. The speed of a proton can be found using the fundamental equation for magnetic force: \[F = qvB \sin \theta\] Here, each symbol represents a crucial element:

• \( F \) is the magnetic force exerted on the proton.
• \( q \) is the charge of the proton, which is a constant \(1.60 \times 10^{-19} \mathrm{~C}\).
• \( v \) is the speed we aim to find.
• \( B \) is the magnetic field strength.
• \( \theta \) is the angle between the proton's velocity and the magnetic field direction.

To isolate the proton's speed \( v \), rearrange the formula: \[v = \frac{F}{qB \sin \theta}\] This equation shows the speed depends on the force, the proton's charge, the field strength, and the angle of entry. Calculating \( \sin(23.0^{\circ}) \approx 0.3907 \) lets us find \( v \), which turns out to be approximately \( 1.073 \times 10^7 \mathrm{~m/s} \). This speed indicates how fast the proton moves within the magnetic field.

Kinetic Energy

Once the proton's speed is determined, we can calculate its kinetic energy. Kinetic energy (KE) represents the energy that a particle possesses due to its motion. The formula to compute kinetic energy is:\[KE = \frac{1}{2} mv^2\]

• \( m \) is the mass of the proton, usually \( 1.67 \times 10^{-27} \mathrm{~kg} \).
• \( v \) is the speed determined previously.

Substituting the values, the kinetic energy in joules can be calculated. This calculated energy can also be expressed in electron-volts (eV), another common energy unit in physics. To convert the kinetic energy into electron-volts, use:\[KE\text{ in eV} = \frac{KE\text{ in joules}}{1.60 \times 10^{-19}}\]This conversion is useful since eV is a typical unit for small-scale energies like those in atomic physics.

Magnetic Field Strength

Magnetic field strength, denoted as \( B \), is a crucial factor in influencing how a charged particle like a proton behaves in a magnetic field. The field strength determines the magnitude of the force applied to moving charges.

• The magnetic field strength in this exercise is given as \(2.60 \mathrm{mT}\), which you must convert into teslas for calculations, hence \(2.60 \times 10^{-3} \mathrm{~T}\).
• Magnetic field strength affects the amount of curvature in the path of the charged particle. Stronger fields result in more significant force, altering the proton's trajectory more extensively.
• The force is also directly linked to the angle \( \theta \). At \( \theta = 0^{\circ} \) or \(180^{\circ} \), where \( \sin \theta = 0 \), no force acts on the proton. Conversely, when the angle is \(90^{\circ}\), the force is maximized, showcasing the directional dependence.

This concept illuminates how magnetic forces and field strengths interplay to define paths and energies of particles within fields, a foundational idea in electromagnetism.