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A proton is projected perpendicularly into a magnetic field that has a magnitude of \(0.50 \mathrm{~T}\). The field is then adjusted so that an electron will follow the exact same circular path when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Verify that your answer is consistent with your answers to the Concept Questions.

Short Answer

Expert verified
The magnitude of the field for the electron is approximately 0.00137 T.

Step by step solution

01

Understand the Problem

We are given that a proton and an electron both follow the same circular path when projected perpendicularly into a magnetic field. The fields are different for the two particles, but they have the same speed when entering the respective fields. We need to find the magnitude of the magnetic field for the electron that allows it to follow the same path as the proton.
02

Recall the Formula for Circular Motion in a Magnetic Field

For a charged particle moving in a magnetic field, the centripetal force is provided by the magnetic force. The equation for this condition is:\[ qvB = \frac{mv^2}{r} \]where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field, \( m \) is the mass, and \( r \) is the radius of the circular path. The equation simplifies to:\[ r = \frac{mv}{qB} \]
03

Set Equal Radii for Proton and Electron

Since the radius of the circular path is the same for both the proton and electron, set the radii equal from the formula:\[ \frac{m_p v}{q_p B_p} = \frac{m_e v}{q_e B_e} \]where \( m_p \) and \( q_p \) are the mass and charge of the proton, and \( m_e \) and \( q_e \) are the mass and charge of the electron. \( B_p \) is given as 0.5 T.
04

Solve for Electron's Magnetic Field, \( B_e \)

Since the charge of the proton \( q_p = +e \) and the charge of the electron \( q_e = -e \), the absolute values are equal: \( |q_p| = |q_e| = e \). Solve for \( B_e \):\[ \frac{m_p v}{e B_p} = \frac{m_e v}{e B_e} \]Cancel the common terms (e, v):\[ \frac{m_p}{B_p} = \frac{m_e}{B_e} \]Rearrange to find \( B_e \):\[ B_e = \frac{m_e}{m_p} B_p \]Use the known masses: \( m_p \approx 1.67 \times 10^{-27} \) kg and \( m_e \approx 9.11 \times 10^{-31} \) kg, and \( B_p = 0.5 \) T.
05

Calculate the Magnitude of the Magnetic Field for Electron

Substitute the known values into the equation for \( B_e \):\[ B_e = \frac{9.11 \times 10^{-31}}{1.67 \times 10^{-27}} \times 0.50 \]\[ B_e \approx \frac{9.11}{1670} \times 0.50 \]\[ B_e \approx 0.00273 \times 0.50 \]\[ B_e \approx 0.001365 \] T.Therefore, the magnitude of the field for the electron is approximately 0.00137 T.

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Key Concepts

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

Circular Motion
When charged particles like protons and electrons move perpendicular to a magnetic field, they experience a force that causes them to move in a circular path. This kind of motion is known as circular motion. In circular motion, the magnetic force acts as the centripetal force that keeps the particle moving along the curved path. The balance between these forces can be expressed mathematically by the equation: \[ qvB = \frac{mv^2}{r} \]where:
  • \( q \) is the charge of the particle,
  • \( v \) is its velocity,
  • \( B \) is the magnetic field strength,
  • \( m \) is the particle mass, and
  • \( r \) is the radius of the circle.
To better understand the motion in a magnetic field, it is crucial to realize that the particle continues moving in circles as long as the velocity remains perpendicular to the magnetic field. If the velocity of the particle changes or becomes parallel to the field, the nature of its path will change.
Charged Particles
Charged particles are particles with either a positive or negative electric charge. Protons and electrons are typical examples of such particles. The behavior of these particles in a magnetic field is influenced by their charge and mass.

For a particle to experience a magnetic force, it must be charged. The formula governing the circular motion of a charged particle in a magnetic field incorporates the charge \( q \). If the charge is positive, like a proton, the particle's direction will be affected differently compared to a negatively charged particle, like an electron.

The charge of the particle also affects the strength and direction of the magnetic force experienced. Essentially, this means the trajectory, including the radius of the circular path, can differ even for particles entering the field at the same velocity, just like in the exercise provided.
Proton and Electron
Protons and electrons are fundamental particles with distinct properties but are both crucial in understanding magnetic fields.

**Protons:**
  • They have a positive charge \( +e \), where \( e \approx 1.6 \times 10^{-19} \) Coulombs.
  • Their mass is approximately \( 1.67 \times 10^{-27} \) kg.
**Electrons:**
  • They carry a negative charge \( -e \), which in absolute magnitude is equal to the charge of a proton.
  • Their mass is much smaller, about \( 9.11 \times 10^{-31} \) kg.
Despite having the same magnitude of charge, the difference in their masses plays a crucial role. When both are subject to the same magnetic field, this difference impacts the magnetic force they experience and their subsequent motion paths in that field. Thus, when finding equivalent paths for both, equations must be adjusted to accommodate their mass variances.
Magnetic Force
The magnetic force acts perpendicular to both the magnetic field and the velocity of a charged particle moving through it. This force is the reason charged particles move in a curved or circular path within a magnetic field.

The magnitude of this force is defined by the equation:\[ F = qvB \sin \theta \]where \( q \) is the charge, \( v \) is the velocity of the particle, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field. For perpendicular motion, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \). Hence, the forceis at its maximum value: \( F = qvB \).

This force plays a key role in how the proton and electron in our exercise can follow the same path in different magnetic fields. Since they have the same charge magnitude but differing masses, their magnetic force experiences adjust according to their respective masses, requiring changes in the magnetic field strength to keep their paths identical.

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Most popular questions from this chapter

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{~T},\) and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{~A} .\) What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

A particle that has an \(8.2-\mu C\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). When the velocity points along the \(+x\) axis, the particle experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{~N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field. Make sure that your answers are consistent with your answers to the Concept Questions.

Consult Interactive Solution \(\underline{21.43} 21.43\) at to see how this problem can be solved. The coil in Figure \(21-22 a\) contains 410 turns and has an area per turn of \(3.1 \times 10^{-3} \mathrm{~m}^{2}\). The magnetic field is \(0.23 \mathrm{~T},\) and the current in the coil is \(0.26 \mathrm{~A} .\) A brake shoe is pressed perpendicularly against the shaft to keep the coil from turning. The coefficient of static friction between the shaft and the brake shoe is \(0.76 .\) The radius of the shaft is \(0.012 \mathrm{~m}\). What is the magnitude of the minimum normal force that the brake shoe exerts on the shaft?

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{~m} / \mathrm{s}\) through a 3.6-T magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{~N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{~m}^{2}\). What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

Two isotopes of carbon, carbon- 12 and carbon- \(13,\) have masses of \(19.93 \times 10^{-27} \mathrm{~kg}\) and \(21.59 \times 10^{-27} \mathrm{~kg},\) respectively. These two isotopes are singly ionized \((+e)\) and each is given a speed of \(6.667 \times 10^{5} \mathrm{~m} / \mathrm{s}\). The ions then enter the bending region of a mass spectrometer where the magnetic field is \(0.8500 \mathrm{~T}\). Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

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