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An electron that has velocity $$ \vec{v}=\left(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ moves through the uniform magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-\) \((0.15 \mathrm{~T}) \hat{\mathrm{j}}\). (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
The force on the electron is \(6.24 \times 10^{-14}\,\hat{k}\) N, and on the proton is \(-6.24 \times 10^{-14}\,\hat{k}\) N.

Step by step solution

01

Understand the magnetic force equation

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

Identify charges for electron and proton

The charge of an electron is \( q_e = -1.6 \times 10^{-19} \text{ C} \) and the charge of a proton is \( q_p = 1.6 \times 10^{-19} \text{ C} \).
03

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

The velocity vector is \( \vec{v} = (2.0 \times 10^6 \hat{i} + 3.0 \times 10^6 \hat{j}) \text{ m/s} \) and the magnetic field vector is \( \vec{B} = (0.030 \hat{i} - 0.15 \hat{j}) \text{ T} \). Compute the cross product:\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \2.0 \times 10^6 & 3.0 \times 10^6 & 0 \0.030 & -0.15 & 0\end{vmatrix}\]After calculation, this results in:\( \vec{v} \times \vec{B} = (0 \hat{i} + 0 \hat{j} + (-0.3 - 0.09) \hat{k}) \times 10^6 \text{ m/s \ T} = -3.9 \times 10^5 \hat{k} \text{ m/s \ T} \).
04

Calculate force on the electron

Using \( \vec{F} = q(\vec{v} \times \vec{B}) \) with the electron's charge:\[ \vec{F}_e = (-1.6 \times 10^{-19}\,\text{C})(-3.9 \times 10^{5}\,\hat{k}\,\text{N}) = 6.24 \times 10^{-14}\,\hat{k}\,\text{N} \].
05

Calculate force on the proton

Using \( \vec{F} = q(\vec{v} \times \vec{B}) \) with the proton's charge:\[ \vec{F}_p = (1.6 \times 10^{-19}\,\text{C})(-3.9 \times 10^{5}\,\hat{k}\,\text{N}) = -6.24 \times 10^{-14}\,\hat{k}\,\text{N} \].

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Key Concepts

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

Lorentz Force
The Lorentz force is fundamental in understanding how charged particles interact with magnetic fields. It is expressed as \( \vec{F} = q(\vec{v} \times \vec{B}) \). Here, \( q \) represents the charge of the particle, \( \vec{v} \) is the velocity at which it travels, and \( \vec{B} \) stands for the magnetic field vector. This equation tells us that the force on the particle not only depends on these variables but is also the result of a cross product, which means it will always be perpendicular to both \( \vec{v} \) and \( \vec{B} \). This perpendicular nature is why charged particles perform spiral or circular motions in magnetic fields.
Cross Product
Calculating the cross product is key in determining the Lorentz force. The cross product \( \vec{v} \times \vec{B} \) is calculated using a determinant:+ When you set up the determinant with the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and fill in the velocity \( (2.0 \times 10^6 \hat{i} + 3.0 \times 10^6 \hat{j}) \) and the magnetic field \( (0.030 \hat{i} - 0.15 \hat{j}) \), you get specific components.+ This results in the vector \( (0 \hat{i} + 0 \hat{j} - 3.9 \times 10^5 \hat{k}) \), which signifies the direction of the magnetic force relative to the motion and field.Calculating this correctly is crucial because it tells us in which direction the velocity of the particle will alter as it travels through the magnetic field.
Charge of Electron and Proton
The charge of an electron is \( -1.6 \times 10^{-19} \) C, while a proton carries a charge of \( 1.6 \times 10^{-19} \) C. Although they have opposite signs, the magnitude of their charges is the same, which means that they will experience equal magnitude of force in a given magnetic field if their velocities are identical. However, due to the negative charge, the force on the electron will be in the opposite direction compared to the proton. Understanding these charges' effects helps predict the behaviors of electron and proton in the same magnetic environment, such as their paths diverging based on their charge type when subjected to the same forces.
Magnetic Field
A magnetic field is a vector field that exerts force on particles that are in motion. It is represented as \( \vec{B} \) and measured in teslas (T). The direction of the field affects the direction of the force applied to a moving charge. In the given case, the field is expressed by \( (0.030 \hat{i} - 0.15 \hat{j}) \) T, showing it has components along \( \hat{i} \) and \( \hat{j} \) axes. The minus sign indicates the field's direction along the \( \hat{j} \)-axis is opposite to the conventional positive direction. When a charged particle such as an electron or proton moves through this field, the magnetic force calculated takes these vector components into account, determining the eventual path and behavior of the particle.

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Most popular questions from this chapter

A particle with charge \(2.0 \mathrm{C}\) moves through a uniform magnetic field. At one instant the velocity of the particle is \((2.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}+6.0 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\) and the magnetic force on the particle is \((4.0 \hat{\mathrm{i}}-20 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}) \mathrm{N}\). The \(x\) and \(y\) components of the magnetic field are equal. What is \(\vec{B}\) ?

A proton, a deuteron \((q=+e, m=2.0 \mathrm{u})\), and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) all having the same kinetic energy enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). What is the ratio of (a) the radius \(r_{d}\) of the deuteron path to the radius \(r_{p}\) of the proton path and (b) the radius \(r_{\alpha}\) of the alpha particle path to \(r_{2} ?\)

A horizontal power line carries a current of \(5000 \mathrm{~A}\) from south to north. Earth's magnetic field \((60.0 \mu \mathrm{T})\) is directed toward the north and inclined downward at \(70.0^{\circ}\) to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on \(100 \mathrm{~m}\) of the line due to Earth's field.

A proton travels through uniform magnetic and electric fields. The magnetic field is \(\vec{B}=-2.50 \hat{\mathrm{i}} \mathrm{mT}\). At one instant the velocity of the proton is \(\vec{v}=2000 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s} .\) At that instant and in unit-vector notation, what is the net force acting on the proton if the electric field is (a) \(4.00 \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}\), (b) \(-4.00 \hat{\mathrm{k}} \mathrm{V} / \mathrm{m}\), and \((\mathrm{c})\) \(4.00 \hat{\mathrm{i}} \mathrm{V} / \mathrm{m} ?\)

A magnetic dipole with a dipole moment of magnitude \(0.020 \mathrm{~J} / \mathrm{T}\) is released from rest in a uniform magnetic field of magnitude \(52 \mathrm{~m} \mathrm{~T}\). The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is \(0.80 \mathrm{~mJ}\). (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

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