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Chapter 27: Problem 9

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 \(\mathrm{km} / \mathrm{s}\) in the \(+x\) -direction experiences a force of \(2.25 \times 10^{-16} \mathrm{N}\) in the \(+y\) -direction, and an electron moving at 4.75 \(\mathrm{km} / \mathrm{s}\) in the \(-z\) -direction experiences a force of \(8.50 \times\) \(10^{-16} \mathrm{N}\) in the \(+y\) -direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\) -direction at 3.20 \(\mathrm{km} / \mathrm{s}\) ?

Short Answer

Expert verified
Magnetic field is 0.9375 T in the +z-direction. Magnetic force on the electron is 4.80 x 10^{-16} N in the -x-direction.

Step by step solution

01

Understanding the Lorentz Force

The magnetic force experienced by a charged particle moving in a magnetic field is given by the Lorentz force equation: \( F = q(v \times B) \), where \( q \) is the charge of the particle, \( v \) is the velocity vector, and \( B \) is the magnetic field vector.
02

Force on the Proton

A proton experiences a force of \( 2.25 \times 10^{-16} \, \text{N} \) in the \( +y \)-direction when moving at \( 1.50 \, \text{km/s} \) in the \( +x \)-direction. The charge of a proton is \( q_p = 1.6 \times 10^{-19} \, \text{C} \). Using the Lorentz force formula, the equation becomes \( 2.25 \times 10^{-16} = 1.6 \times 10^{-19} \cdot 1500 \cdot B \).
03

Solving for Magnetic Field from Proton's Data

We solve for \( B \): \[ B = \frac{2.25 \times 10^{-16}}{1.6 \times 10^{-19} \times 1500} \approx 0.9375 \, \text{T} \].
04

Component Direction Confirmation using Electron

An electron moving at \(4.75 \, \text{km/s}\) in the \(-z\)-direction experiences a force \(8.50 \times 10^{-16} \, \text{N}\) in the \(+y\)-direction. The charge of an electron is \( q_e = -1.6 \times 10^{-19} \, \text{C} \). Using \( F = q(v \times B) \), and considering the negative charge: \[8.50 \times 10^{-16} = (-1.6 \times 10^{-19})(-4750)B \]. Solving confirms \( B \) and dictates the z-component of \(v \times B\) is \(+y\), confirming \( B = 0.9375 \, \text{T} \, (+z)\).
05

Solve for Force on Electron in -y Direction

To find the force on an electron moving in the \(-y\)-direction, use \( F = q(v \times B) \), \( v = 3200 \times 10^3 \, \text{m/s} \), and \( q = -1.6 \times 10^{-19} \). The resultant force direction will be in the \(-x\)-direction. Magnitude is \[ F = |(-1.6 \times 10^{-19})(-3200 \times 0.9375)| \]. Calculating gives: \[ F = 4.80 \times 10^{-16} \, \text{N}\].

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Key Concepts

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

Magnetic Field Direction
Understanding the direction of a magnetic field is crucial when analyzing the motion of charged particles. According to the right-hand rule, if you point your thumb in the direction of a positively charged particle's velocity and your fingers in the direction of the magnetic force experienced by the particle, your palm faces the direction of the magnetic field.

In the given problem, a proton moves in the \(+x\)-direction and experiences a force in the \(+y\)-direction. This implies that the magnetic field is directed along the \(+z\)-axis. For other particles, like electrons, similar reasoning applies, keeping in mind their negative charge which results in a magnetic force in the opposite direction.
Charged Particles Motion
The motion of charged particles in a magnetic field is profoundly influenced by the Lorentz force. This force is perpendicular to both the velocity of the particle and the magnetic field, causing charged particles to move in a circular or helical path depending on their initial velocity and the uniformity of the magnetic field.

In our context, when a proton or an electron enters a magnetic field, it doesn鈥檛 move in a straight line but curves based on the field's direction. The curvature of the path depends on the particle鈥檚 charge, speed, and the magnetic field strength. For instance, the movement of our proton and electron in the exercise confirms these principles.
Magnetic Force Calculation
Magnetic force calculation is done using the Lorentz force equation: \( F = q(v \times B) \).Here, the force depends on:
  • \( q \), the charge of the particle.
  • \( v \), the velocity of the particle.
  • \( B \), the magnetic field.

In the exercise, calculations were performed for both a proton and an electron.
For instance, the force on a proton was calculated by rearranging the formula to find the magnetic field strength, knowing the force and velocity.
Thus, \( B = \frac{F}{q \cdot v} = \frac{2.25 \times 10^{-16}}{1.6 \times 10^{-19} \cdot 1500} \), resulting in a field strength of approximately 0.9375 T. This consistency with the electron鈥檚 data confirmed the field's magnitude and direction.
Proton and Electron Charge
The charge of subatomic particles plays a significant role in calculating magnetic forces. Protons, with a positive charge of \( +1.6 \times 10^{-19} \, \text{C} \), and electrons, with a negative charge of \( -1.6 \times 10^{-19} \, \text{C} \), experience forces in opposite directions under identical conditions.

In the given exercise, these charges were pivotal in determining the forces acting on them when they moved through the magnetic field. The positive charge of a proton leads to a force in the expected direction, while an electron, due to its negative charge, experiences a force in the opposite direction relative to the conventional right-hand rule predictions.

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Most popular questions from this chapter

A particle with charge \(6.40 \times 10^{-19} \mathrm{C}\) travels in a circular orbit with radius 4.68 \(\mathrm{mm}\) due to the force exerted on it by a magnetic field with magnitude 1.65 \(\mathrm{T}\) and perpendicular to the orbit. (a) What is the magnitude of the linear momentum \(\vec{p}\) of the particle? (b) What is the magnitude of the angular momentum \(\vec{L}\) of the particle?

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\vec{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\imath}+(7.40 \times\) \(10^{-7} \mathrm{N} ) \hat{J}\) . (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v} \cdot\) What is the angle between \(\vec{v}\) and \(\vec{F} ?\)

A \(150-\) g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 \(\mathrm{T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

An electron in the beam of a TV picture tube is accelerated by a potential difference of 2.00 \(\mathrm{kV}\) . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 \(\mathrm{m} .\) What is the magnitude of the field?

You wish to hit a target from several meters away with a charged coin having a mass of 4.25\(\mathrm { g }\) and a charge of \(+ 2500 \mu \mathrm { C }\) . The coin is given an initial velocity of 12.8\(\mathrm { m } / \mathrm { s }\) , and a downward, uniform electric field with field strength 27.5\(\mathrm { N } / \mathrm { C }\) exists through-out the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target?
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