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Chapter 29: Problem 3

Proton Traveling 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.

Short Answer

Expert verified
(a) 2.58 \times 10^{6} \mathrm{~m/s}(b)3.48 \times 10^{4} \mathrm{~eV}

Step by step solution

01

Understand the Formula for Magnetic Force

The formula for the magnetic force experienced by a charge moving in a magnetic field is given by \( F = qvB \sin\theta \) where: \( F \) is the magnetic force \( q \) is the charge of the proton \( v \) is the velocity of the proton \( B \) is the magnetic field strength \( \theta \) is the angle between the proton's velocity vector and the magnetic field.
02

Rearrange the Formula to Solve for Velocity (v)

Rearrange the magnetic force formula to solve for the proton's velocity (\( v \)). \( v = \frac{F}{qB \sin\theta} \)
03

Substitute Given Values and Solve for the Proton's Speed

Substitute the given values into the rearranged formula to find the proton's speed. Given: \( F = 6.50 \times 10^{-17} \mathrm{~N} \) \( q = 1.60 \times 10^{-19} \mathrm{~C} \) \( B = 2.60 \times 10^{-3} \mathrm{~T} \) \( \theta = 23.0^{\circ} \) \( \sin 23.0^{\circ} = 0.3907 \) Now substituting the values, we get \( v = \frac{6.50 \times 10^{-17}}{(1.60 \times 10^{-19}) (2.60 \times 10^{-3}) (0.3907)} \approx 2.58 \times 10^{6} \mathrm{~m/s} \)
04

Formula for Kinetic Energy

The kinetic energy (\(KE\)) of a particle is given by: \( KE = \frac{1}{2} mv^2 \) where: \(m\) is the mass of the proton (\( 1.67 \times 10^{-27} \mathrm{~kg} \) and \(v\) is the speed of the proton.
05

Convert Kinetic Energy to Electron-volts

First, calculate the kinetic energy in joules and then convert it to electron-volts. Given: \( v = 2.58 \times 10^6 \mathrm{~m/s} \) \( m = 1.67 \times 10^{-27} \mathrm{~kg} \) \( KE = \frac{1}{2} (1.67 \times 10^{-27}) (2.58 \times 10^6)^2 \approx5.56 \times 10^{-15} \mathrm{ ~J} \) To convert joules to electron-volts, use the conversion factor: \( 1 \text{ eV} = 1.60 \times 10^{-19} \text{ J} \) Thus, the kinetic energy in electron-volts is: \( \frac{5.56 \times 10^{-15}}{1.60 \times 10^{-19}} \approx 3.48 \times 10^{4} \text{ eV} \)

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Key Concepts

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

Magnetic Force
When a charged particle, like a proton, moves through a magnetic field, it experiences a magnetic force. This force is described by the equation: \( F = qvB \, \sin\theta \)where:
  • \( F \) is the magnetic force
  • \( q \) is the charge of the proton, approximately \(1.60 \times 10^{-19} \mathrm{~C}\)
  • \( v \) is the velocity of the proton
  • \( B \) is the magnetic field strength
  • \( \theta \) is the angle between the proton's velocity and the magnetic field
In our problem,
  • \( F = 6.50 \times 10^{-17} \mathrm{~N}\)
  • \( B = 2.60 \times 10^{-3} \mathrm{~T}\)
  • \( \theta = 23.0^{\circ} \)
Using the given data, we can find the proton's velocity.
Velocity Calculation
To calculate the proton's speed from the magnetic force formula, we rearrange it to solve for \( v \):\( v = \frac{F}{qB \, \sin\theta} \)Plugging in our values:
  • \( F = 6.50 \times 10^{-17} \mathrm{~N}\)
  • \( q = 1.60 \times 10^{-19} \mathrm{~C}\)
  • \( B = 2.60 \times 10^{-3} \mathrm{~T}\)
  • \( \sin 23.0^{\circ} = 0.3907 \)
We substitute these into the formula:\( v = \frac{6.50 \times 10^{-17}}{(1.60 \times 10^{-19}) (2.60 \times 10^{-3}) (0.3907)} \approx 2.58 \times 10^{6} \mathrm{~m/s} \)So, the proton's speed is approximately \(2.58 \times 10^{6} \mathrm{~m/s}\).
Kinetic Energy
The kinetic energy (KE) of a moving particle can be calculated using the formula:\( KE = \frac{1}{2} mv^2 \)where:
  • \( m \) is the mass of the particle
  • \( v \) is the velocity
For a proton:
  • \( m \approx 1.67 \times 10^{-27} \mathrm{~kg} \)
  • \( v \approx 2.58 \times 10^{6} \mathrm{~m/s} \)
Substituting these into the kinetic energy formula:\( KE = \frac{1}{2} (1.67 \times 10^{-27}) (2.58 \times 10^6)^2 \approx 5.56 \times 10^{-15} \mathrm{~J} \)Thus, the proton's kinetic energy is approximately \(5.56 \times 10^{-15} \mathrm{~J}\).
Electron-Volts
To convert kinetic energy from joules to electron-volts (eV), we use the conversion factor: 1 electron-volt = \( 1.60 \times 10^{-19} \mathrm{~J} \). Using this factor, we convert the kinetic energy found earlier:\( \text{KE} = \frac{5.56 \times 10^{-15} \text{~J}}{1.60 \times 10^{-19} \text{~J/eV}} \approx 3.48 \times 10^{4} \text{~eV} \)So, the proton's kinetic energy in electron-volts is approximately \(3.48 \times 10^{4} \text{~eV} \). Understanding electron-volts is important because it provides a more intuitive scale for measuring the energy of particles at the atomic and subatomic levels.

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Most popular questions from this chapter

Current Loop A single-turn current loop, carrying a current of \(4.00 \mathrm{~A}\), is in the shape of a right triangle with sides \(50.0,120\), and \(130 \mathrm{~cm}\). The loop is in a uniform magnetic field of magnitude \(75.0 \mathrm{mT}\) whose direction is parallel to the current in the \(130 \mathrm{~cm}\) side of the loop. (a) Find the magnitude of the magnetic force on each of the three sides of the loop. (b) Show that the total magnetic force on the loop is zero.

Along the \(x\) Axis A wire \(50 \mathrm{~cm}\) long lying along the \(x\) axis carries a current of \(0.50 \mathrm{~A}\) in the positive \(x\) direction. It passes through a magnetic field \(\vec{B}=(0.0030 \mathrm{~T}) \hat{\mathrm{j}}+(0.0100 \mathrm{~T}) \hat{\mathrm{k}} .\) Find the magnetic force on the wire.

Horizontal Motion An electron with kinetic energy \(2.5 \mathrm{keV}\) moves horizontally into a region of space in which there is a downward-directed uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\). (a) What are the magnitude and direction of the (smallest) uniform magnetic field that will cause the electron to continue to move horizontally? Ignore the gravitational force, which is small. (b) Is it possible for a proton to pass through the combination of fields undeflected? If so, under what circumstances?

Source of Electrons A source injects an electron of speed \(v=\) \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\) into a uniform magnetic field of magnitude \(B=\) \(1.0 \times 10^{-3} \mathrm{~T}\). The velocity of the electron makes an angle \(\theta=10^{\circ}\) with the direction of the magnetic field. Find the distance \(d\) from the point of injection at which the electron next crosses the field line that passes through the injection point.

Proton, Deuteron, Alpha A proton, a deuteron \((q=+e, m=\) \(2.0 \mathrm{u})\), and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) with the same kinetic energies enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). Compare the radii of their circular paths.
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