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A proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=(-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r,\) of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f,\) of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

Short Answer

Expert verified
Question: A proton moves at a speed of \(1.00 \cdot 10^{6}\,\mathrm{m/s}\) at an angle of \(60^{\circ}\) with respect to the positive \(z\)-axis. If the magnetic field is pointing in the \(-z\) direction and has a magnitude of \(0.500\,\mathrm{T}\), determine the following: (a) Describe the trajectory of the proton. (b) Calculate the radius of the trajectory projected onto the XY-plane. (c) Calculate the period and frequency of the motion in the XY-plane. (d) Calculate the pitch of the motion. Answer: (a) The trajectory of the proton is helical. (b) The radius of the trajectory projected onto the XY-plane is \(0.179\,\mathrm{m}\). (c) The period of the motion in the XY-plane is \(6.58 \cdot 10^{-8}\,\mathrm{s}\), and the frequency is \(1.52 \cdot 10^{7}\,\mathrm{Hz}\). (d) The pitch of the motion is \(3.29 \cdot 10^{-2}\,\mathrm{m}\).

Step by step solution

01

Analyze the motion of the proton in a magnetic field

The motion of a charged particle in a magnetic field depends on the velocity of the particle and the direction of the magnetic field. If the velocity vector is parallel or anti-parallel to the magnetic field, the particle moves in a straight line. Otherwise, the particle moves in a helical path. In this case, the velocity vector of the proton makes an angle of \(60^{\circ}\) with the positive \(z\)-axis, and the magnetic field points in the opposite direction, \(-z\). So, the motion of the proton would be helical.
02

Calculate the radius of the trajectory projected onto the XY-plane

The radius \(r\) of the trajectory projected onto the XY-plane can be calculated using the formula for the radius of the circular motion in a magnetic field, \(r = \frac{mv_\perp}{qB}\) where \(m\) is the mass of the proton, \(v_\perp\) is the component of the velocity perpendicular to the magnetic field, \(q\) is the charge of the proton, and \(B\) is the magnitude of the magnetic field. First, let's calculate \(v_\perp\). Since the velocity vector makes an \(60^{\circ}\) angle with the positive \(z\)-axis, the component of the velocity in the XY-plane is given by, \(v_\perp = v\sin{\theta} = (1.00 \cdot 10^{6}\,\mathrm{m/s})\sin{(60^{\circ})} = 8.66 \cdot 10^{5}\,\mathrm{m/s}\) Now, let's find the radius \(r\). The proton's mass is \(1.67 \cdot 10^{-27} \mathrm{kg}\), its charge is \(1.60 \cdot 10^{-19} \mathrm{C}\), and the magnitude of the magnetic field is \(0.500\,\mathrm{T}\). Using these values, we get, \(r = \frac{(1.67 \cdot 10^{-27}\,\mathrm{kg})(8.66 \cdot 10^{5}\,\mathrm{m/s})}{(1.60 \cdot 10^{-19}\,\mathrm{C})(0.500\,\mathrm{T})} = 0.179\,\mathrm{m}\)
03

Calculate the period and frequency of the motion in the XY-plane

The period \(T\) of the circular motion in a magnetic field can be calculated using the formula, \(T = \frac{2\pi m}{qB}\) The frequency \(f\) of the proton's motion can be calculated as the reciprocal of the period, \(f = \frac{1}{T}\) Using the known values, we get, \(T = \frac{2\pi (1.67 \cdot 10^{-27}\,\mathrm{kg})}{(1.60 \cdot 10^{-19}\,\mathrm{C})(0.500\,\mathrm{T})} = 6.58 \cdot 10^{-8}\,\mathrm{s}\) \(f = \frac{1}{6.58 \cdot 10^{-8}\,\mathrm{s}} = 1.52 \cdot 10^{7}\,\mathrm{Hz}\)
04

Calculate the pitch of the motion

The pitch is the distance traveled by the proton in the direction of the magnetic field in one period. Since the velocity vector has a component along the magnetic field direction, \(v_{\parallel} = v\cos{\theta}\), the pitch \(P\) can be calculated as, \(P = v_{\parallel}T\) Using the known values, we get, \(P = (1.00 \cdot 10^{6}\,\mathrm{m/s})\cos{(60^{\circ})}(6.58 \cdot 10^{-8}\,\mathrm{s}) = 3.29 \cdot 10^{-2}\,\mathrm{m}\) So, the pitch of the helical motion of the proton in the given magnetic field is \(3.29 \cdot 10^{-2}\,\mathrm{m}\).

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Key Concepts

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

Helical Trajectory
When a charged particle such as a proton moves through a magnetic field, it can take on a unique spiraling motion known as a helical trajectory. This occurs when the particle's velocity is at an angle to the magnetic field rather than directly parallel or anti-parallel. In our example, the proton's velocity vector forms a 60-degree angle with the z-axis, and the magnetic field is directed along the negative z-axis. As a result, the proton spirals around the magnetic field lines while simultaneously moving forward along the z-axis, creating a 3D spring-like path.

This motion is a combination of two components: circular motion in the plane perpendicular to the field lines (in the xy-plane) and linear motion parallel to the field lines. Understanding the helical trajectory of charged particles is crucial for fields such as plasma physics and the study of cosmic rays, as well as in designing devices like cyclotrons and mass spectrometers.
Radius of Circular Motion
The radius of circular motion for a charged particle in a magnetic field is one of the fundamental quantities describing the helical path. It is the distance from the center of the circular path, projected onto the plane perpendicular to the magnetic field, to the particle's position. This radius can be determined using the formula \(r = \frac{mv_{\perp}}{qB}\) where \(m\) is the mass of the particle, \(v_{\perp}\) is its perpendicular velocity component, \(q\) is its charge, and \(B\) is the magnetic field's magnitude.

For our proton, the perpendicular component of its velocity was calculated to be \(8.66 \times 10^{5} \mathrm{m/s}\) since the velocity vector makes a 60-degree angle with the magnetic field. From this information, the radius of the proton's circular motion was determined to be \(0.179 \mathrm{m}\). This figure is key for understanding the scale of the proton's helical motion in relation to the magnetic field.
Period and Frequency
The period and frequency are critical characteristics of any cyclical motion, like the circular component of the proton's helical trajectory. The period \(T\) represents the time it takes for one complete cycle of motion. In the context of a charged particle in a magnetic field, the period is \( T = \frac{2\pi m}{qB} \), dependent on the particle's mass \(m\), charge \(q\), and the magnetic field's strength \(B\). For our scenario, the proton's period is roughly \(6.58 \times 10^{-8} \mathrm{s}\).

The frequency \(f\), on the other hand, is the reciprocal of the period and indicates how many cycles occur in one second. We calculated the proton's frequency to be \(1.52 \times 10^{7} \mathrm{Hz}\), meaning it completes around 15 million cycles per second. Both period and frequency are essential for predicting the behavior of charged particles in various applications, from medical imaging to particle accelerators.
Magnetic Pitch
The pitch of a helical trajectory reflects how far a particle moves along the direction of the magnetic field for every full cycle of its circular motion. Think of it as the 'forward progress' for each spiral turn. It's given by the product of the component of the velocity parallel to the magnetic field \(v_{\parallel}\) and the period \(T\). Mathematically, \(P = v_{\parallel}T\).

With the proton's velocity parallel to the magnetic field being half of its total velocity due to the 60-degree angle, and considering the previously calculated period, the resultant pitch is approximately \(3.29 \times 10^{-2} \mathrm{m}\), or about 3.29 centimeters. This concept helps in visualizing the 'tightness' or 'looseness' of the helical path and is particularly useful when analyzing the motion of charged particles in controlled environments like magnetic traps.

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Most popular questions from this chapter

At \(t=0\) an electron crosses the positive \(y\) -axis \((\) so \(x=0)\) at \(60.0 \mathrm{~cm}\) from the origin with velocity \(2.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. It is in a uniform magnetic field. a) Find the magnitude and the direction of the magnetic field that will cause the electron to cross the \(x\) -axis at \(x=60.0 \mathrm{~cm}\). b) What work is done on the electron during this motion? c) How long will the trip from \(y\) -axis to \(x\) -axis take?

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible?

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.602 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.000 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B}\), acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0},\) of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\) d) The gold ions (represented by the dark gray spheres) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the light gray spheres) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=3.271 \cdot 10^{-25} \mathrm{~kg} .\) Calculate the mass of a molybdenum ion.

A semicircular loop of wire of radius \(R\) is in the \(x y\) -plane, centered about the origin. The wire carries a current, \(i\), counterclockwise around the semicircle, from \(x=-R\) to \(x=+R\) on the \(x\) -axis. A magnetic field, \(\vec{B},\) is pointing out of the plane, in the positive \(z\) -direction. Calculate the net force on the semicircular loop.

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?

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