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Chapter 28: Problem 63

An electron that has an instantaneous velocity of $$ \vec{v}=\left(-5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ is moving through the uniform magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-\) \((0.15 \mathrm{~T}) \hat{j}\). (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.

Short Answer

Expert verified
(a) Electron force: \( -1.056 \times 10^{-13} \hat{k} \, \text{N} \) (b) Proton force: \( 1.056 \times 10^{-13} \hat{k} \, \text{N} \)

Step by step solution

01

Write down the formula for magnetic force

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

Determine the cross product \( \vec{v} \times \vec{B} \)

The cross product of two vectors \( \vec{v} = v_i \hat{i} + v_j \hat{j} \) and \( \vec{B} = B_i \hat{i} + B_j \hat{j} \) can be calculated using:\[ \vec{v} \times \vec{B} = (v_i B_j - v_j B_i) \hat{k}\]Substitute: \(v_i = -5.0 \times 10^6 \, \text{m/s}\), \(v_j = 3.0 \times 10^6 \, \text{m/s}\), \(B_i = 0.030 \, \text{T}\), and \(B_j = -0.15 \, \text{T}\):\[ \vec{v} \times \vec{B} = (-5.0 \times 10^6 \times -0.15 - 3.0 \times 10^6 \times 0.030) \hat{k}\]Simplify:\[ \vec{v} \times \vec{B} = (0.75 \times 10^6 - 0.09 \times 10^6) \hat{k} = 0.66 \times 10^6 \hat{k} \, \text{m}^2/\text{s}^2\]
03

Calculate the force on the electron

Substitute into the formula for magnetic force with the charge of an electron \( q_e = -1.6 \times 10^{-19} \, \text{C}\): \[ \vec{F} = q_e(\vec{v} \times \vec{B}) = (-1.6 \times 10^{-19})(0.66 \times 10^6) \hat{k}\]Calculate:\[ \vec{F} = -1.056 \times 10^{-13} \hat{k} \, \text{N}\]
04

Repeat the calculation for a proton

The proton has the same magnitude of charge but positive, \( q_p = 1.6 \times 10^{-19} \, \text{C}\). Using the same cross product result from Step 2:\[ \vec{F} = q_p(\vec{v} \times \vec{B}) = (1.6 \times 10^{-19})(0.66 \times 10^6) \hat{k}\]Calculate:\[ \vec{F} = 1.056 \times 10^{-13} \hat{k} \, \text{N}\]

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Key Concepts

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

Magnetic Field
A magnetic field is a vector field in which magnetic forces are experienced by moving electric charges or magnetic materials. It is represented by the vector symbol \(\vec{B}\). The strength and direction of a magnetic field linearly influence the force exerted upon charged particles moving within the field. This field is measured in teslas (T).
  • **Directionality:** The direction of a magnetic field is determined with the use of 'right-hand rule', which relates to how the field curls around a current-carrying conductor.
  • **Uniformity:** In this scenario, the magnetic field is described as 'uniform', meaning it has the same strength and direction at every point in space within the area considered.
  • **Effect on Particles:** When a charged particle enters a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field direction.
Understanding magnetic fields is essential in predicting how and where forces will act on particles such as electrons or protons moving through them.
Electron Motion
Electrons are negatively charged particles, and their motion in a magnetic field leads to an interesting phenomenon governed by the Lorentz force law. The direction and magnitude of the force acting on an electron depend on both its velocity and the internal characteristics of the field.
  • **Velocity:** The velocity of an electron is a vector indicating the direction and speed of its motion. In the given exercise, the electron's velocity is specified with components along the \(\hat{i}\) and \(\hat{j}\) directions.
  • **Lorentz Force:** According to the Lorentz Force Law, \(\vec{F} = q(\vec{v} \times \vec{B})\), an electron under a magnetic field experiences a force perpendicular to the direction of its velocity and the magnetic field. For electrons, this cross product is crucial in determining the resultant force direction as they carry a negative charge.
  • **Implications:** When an electron enters a magnetic field, it tends to follow a helical trajectory. The behavior is pivotal in magnetic confinement fusion and designing electron mass spectrometers.
The charge of the electron being negative means the force acts in the opposite direction to that of the cross product outcome.
Proton
Protons are positively charged subatomic particles, and their interactions with magnetic fields are quite similar to those of electrons, except in the direction of the Lorentz force. This occurs due to the positive charge of the proton.
  • **Charge Characteristics:** Protons possess a charge of the same magnitude as electrons (1.6 \(\times\) 10^{-19} C) but are positively charged.
  • **Effect of Magnetic Fields:** When a proton moves in a magnetic field, it experiences a force perpendicular to its path and magnetic field, as governed by the Lorentz force equation \(\vec{F} = q(\vec{v} \times \vec{B})\).
  • **Comparison with Electrons:** Despite moving with the same speed as in the given problem, the force direction observed is opposite to that of an electron due to the opposite charge sign.
  • **Applications:** Protons in magnetic fields are foundational in technologies like cyclotrons, used in particle physics research, and medical treatments through proton beam therapy.
While their charges and masses differ significantly, the principle by which they react to a magnetic field remains consistent with the vector nature of the force involved.
Vector Cross Product
The vector cross product is a mathematical operation applied to two vectors, resulting in a third vector perpendicular to both of the initial vectors. For charged particles in magnetic fields, the cross product helps determine the direction and magnitude of the force applied.
  • **Basics of Cross Product:** For vectors \(\vec{A} = A_i \hat{i} + A_j \hat{j}\) and \(\vec{B} = B_i \hat{i} + B_j \hat{j}\), the cross product \(\vec{A} \times \vec{B} = (A_i B_j - A_j B_i) \hat{k}\) provides a resultant vector along the third perpendicular axis \(\hat{k}\).
  • **Physical Interpretation:** In the context of the Lorentz force, this product indicates how the velocity of a charge, and the magnetic field direction interact to produce a resultant force vector.
  • **Right-Hand Rule:** This rule aids in determining the direction of the vector product: if the fingers follow the velocity and curl towards the magnetic field, the thumb points in the force vector direction.
  • **Application Example:** When referred back to the original problem, finding \(\vec{v} \times \vec{B}\) gives the specific orientation of the force experienced by charged particles like electrons and protons.
The vector nature ensures that forces are calculated not only for magnitude but also for direction, a crucial part of analyzing charged particle dynamics in magnetic fields.

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Most popular questions from this chapter

An electron follows a helical path in a uniform magnetic field given by \(\vec{B}=(20 \hat{\mathrm{i}}-50 \hat{\mathrm{j}}-30 \hat{\mathrm{k}}) \mathrm{mT}\). At time \(t=0\), the electron's velocity is given by \(\vec{v}=(40 \hat{\hat{i}}-30 \hat{j}+50 \hat{k}) \mathrm{m} / \mathrm{s}\). (a) What is the angle \(\phi\) between \(\vec{v}\) and \(\vec{B}\) ? The electron's velocity changes with time. Do (b) its speed and (c) the angle \(\phi\) change with time? (d) What is the radius of the helical path?

A circular wire loop of radius \(15.0 \mathrm{~cm}\) carries a current of \(3.20 \mathrm{~A}\). It is placed so that the normal to its plane makes an angle of \(41.0^{\circ}\) with a uniform magnetic field of magnitude \(12.0 \mathrm{~T}\). (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop?

A long, rigid conductor, lying along an \(x\) axis, carries a current of \(7.0 \mathrm{~A}\) in the negative \(x\) direction. A magnetic field \(\vec{B}\) is present, given by \(\vec{B}=3.0 \mathrm{i}+8.0 x^{2} \hat{\mathrm{j}}\), with \(x\) in meters and \(\vec{B}\) in milliteslas. Find, in unit-vector notation, the force on the \(2.0 \mathrm{~m}\) segment of the conductor that lies between \(x=1.0 \mathrm{~m}\) and \(x=3.0 \mathrm{~m}\).

A conducting rectangular solid of dimensions \(d_{x}=5.00 \mathrm{~m}, d_{y}=3.00 \mathrm{~m}\), and \(d_{z}=2.00 \mathrm{~m}\) moves with a constant velocity \(\vec{v}=(20.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) through a uniform magnetic field \(\vec{B}=(40.0 \mathrm{mT}) \hat{\mathrm{j}}\) (Fig. 28-22). What are the resulting (a) electric field within the solid, in unit- vector notation, and (b) potential difference across the solid? (c) Which face becomes negatively charged?

In Fig. 28-42, an electron with an initial kinetic energy of \(5.0 \mathrm{keV}\) enters region 1 at time \(t=0\). That region contains a uniform magnetic field directed into the page, with magnitude \(0.010 \mathrm{~T}\). The electron goes through a half-circle and then exits region 1, headed toward region 2 across a gap of \(25.0 \mathrm{~cm}\). There is an electric potential difference \(\Delta V=2000 \mathrm{~V}\) across the gap, with a polarity such that the electron's speed increases uniformly as it traverses the gap. Region 2 contains a uniform magnetic field directed out of the page, with magnitude \(0.020 \mathrm{~T}\). The electron goes through a half-circle and then leaves region 2. At what time \(t\) does it leave?
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