91Ó°ÊÓ

A particle with charge \(q\) and mass \(m\) is dropped at time \(t=0\) from rest at its origin in a region of constant magnetic field \(\overrightarrow{\boldsymbol{B}}\) that points horizontally. What happens? To answer, construct a Cartesian coordinate system with the \(y\) -axis pointing downward and the \(z\) -axis pointing in the direction of the magnetic field. At time \(t \geq 0\) the particle has velocity \(\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath} .\) The net force \(\overrightarrow{\boldsymbol{F}}=F_{x} \hat{\imath}+F_{y} \hat{\jmath}\) on the particle is the vector sum of its weight and the magnetic force. (a) Using Newton's second law, write equations for \(a_{x}\) and \(a_{y},\) where \(\vec{a}=a_{x} \hat{\imath}+a_{y} \hat{\jmath}\) is the acceleration of the particle. (b) Differentiate the second of these equations with respect to time. Then substitute your expression for \(a_{x}=d v_{x} / d t\) to determine an equation for \(d v_{y}^{2} / d t^{2}\) in terms of \(v_{y}\) (c) This result shows that \(v_{y}\) is a simple harmonic oscillator. Use the initial conditions to determine \(v_{y}(t) .\) Write your answer in terms of the angular frequency \(\omega=q B / m\). Note that \(\left(d v_{y} / d t\right)_{0}=g .\) (d) Substitute your result for \(v_{y}(t)\) into your equation for \(d v_{x} / d t .\) Integrate using the initial conditions to determine \(v_{x}(t) .\) (e) Integrate your expressions for \(v_{x}(t)\) and \(v_{y}(t)\) to determine \(x(t)\) and \(y(t) .\) (f) If \(m=1.00 \mathrm{mg}, q=19.6 \mu \mathrm{C}\) and \(B=10.0 \mathrm{~T}\), what maximum vertical distance does the particle drop before returning upward?

Step by step solution

01

Form Motion Equations

Use Newton's second law, which states the net force is equal to mass times acceleration, to write the equations of motion. Taking into account the forces' directions and the Lorentz force \( F = q(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}})\), two equations are obtained: \(F_{x}=ma_{x}=q(v_{y}B)\) and \(F_{y}=ma_{y}=mg-q(v_{x}B)\).

02

Differentiate \(a_{y}\) Equation

Differentiate the second equation with respect to time to obtain: \(d^2v_{y}/dt^2 = g - q/m(dv_{x} / dt)B\). Then substitute the expression for \(a_{x}= d v_{x} / dt\) from Step 1 to get \(d^2 v_{y} / dt^{2} = g - v_{y}^2B^2/m\).

03

Solve for \(v_{y}\)

The equation obtained at Step 2 describes a simple harmonic oscillator. The initial condition for \(v_y\) is 0. Hence, its solution is: \(v_{y}(t) = \sqrt{g/ \omega} sin(\omega t)\) with \(\omega = qB/m\). Note: \(\left(d v_{y} / d t\right)_{0}=g\).

04

Solve for \(v_{x}\)

Substitute the \(v_{y}(t)\) obtained in Step 3 into the equation for \(a_{x}=d v_{x} / d t\) and integrate, given the initial condition \(v_x(0)=0\), to get \(v_{x}(t) = - \sqrt{g/ \omega} cos(\omega t)+ \sqrt{g/ \omega}\).

05

Find Position Functions

Integrate \(v_{x}(t)\) and \(v_{y}(t)\) with respect to time to get the functions of position over time. The results are \(x(t)= t\sqrt{g/ \omega}- \sqrt{g/ \omega}sin(\omega t) / \omega\) and \(y(t)=- \sqrt{g/ \omega}cos(\omega t) / \omega\).

06

Compute Maximum Vertical Distance

Given \(m=1.00 \mathrm{mg}, q=19.6 \mu \mathrm{C}\) and \(B=10.0 \mathrm{T}\), compute the maximum vertical distance the particle drops before returning upward. That would be the amplitude of the oscillation, which equals \( \sqrt{g/ \omega}\), or equivalently, \(\sqrt{g \cdot m / (qB)}\).

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Key Concepts

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

Lorentz Force

Imagine a particle with charge zooming through space. When it enters a region with a magnetic field at play, something remarkable happens: the particle starts to experience a force that wasn't there before. This is due to the Lorentz force, a fundamental concept in electromagnetism.

The Lorentz force is the combined effect of electric and magnetic fields on a charged particle. For a particle with charge, denoted as \(q\), moving with a velocity \(\overrightarrow{\boldsymbol{v}}\), in the presence of a magnetic field \(\overrightarrow{\boldsymbol{B}}\), the magnetic part of the Lorentz force is given by the equation \(F = q(\overrightarrow{\boldsymbol{v}} \times \overrightarrow{\boldsymbol{B}})\). In simpler terms, this cross product means that the force is perpendicular to both the velocity of the particle and the magnetic field.

This perpendicular force does not work to speed up or slow down the particle, but instead, it constantly changes the particle's direction, often resulting in a curved path. The Lorentz force is a key player in countless applications, from the paths of cosmic rays and the operation of particle accelerators to everyday gadgets like the magnetic sensors in your smartphone.

Simple Harmonic Oscillator

Now, let's talk about a system that loves to swing back and forth: the simple harmonic oscillator. This is a model that describes a repetitive, or oscillatory, motion, much like a playground swing or a pendulum in a grandfather clock.

Mathematically, a simple harmonic oscillator is any system where the force is directly proportional to the displacement from an equilibrium position and is directed towards that equilibrium position. It's described by the equation \( F = -kx \), where \(k\) is the force constant and \(x\) is the displacement from equilibrium.

In the context of our charged particle in a magnetic field, the particle exhibits simple harmonic motion in the perpendicular direction to the magnetic field when the force is a restoring force – that is, when it acts to bring the particle back to its original position. This results in an oscillation at a specific frequency, known to physicists as the angular frequency \(\omega\). The simple harmonic oscillator is a foundational concept that shows up across many areas of physics, including mechanics, electronics, and quantum mechanics.

Newton's Second Law

At the heart of motion, there's a rule that keeps the universe ticking: Newton's second law. This principle is the engine behind predicting how objects move and accelerate. It states that the force applied to an object is equal to the mass of the object multiplied by its acceleration, which is neatly summed up by the formula \(F = ma\).

For our charged particle, we apply Newton's second law to understand how it accelerates when subject to forces like gravity and the Lorentz force. Separating the motion into components, we can analyze acceleration in the x-direction (\(a_x\)) and y-direction (\(a_y\)). These accelerations are influenced by the magnetic force experienced by the particle, showing us just how interconnected these concepts are.

Using Newton's second law allows us not only to predict the trajectory of our charged particle in a magnetic field but also to understand motion in sports, celestial mechanics, and even the behavior of vehicles and machinery. This law is a cornerstone of classical mechanics and remains an essential tool in physics and engineering.