91Ó°ÊÓ

A particle with negative charge \(q\) and mass \(m=2.58 \times 1\) \(10^{-15} \mathrm{~kg}\) is traveling through a region containing a uniform magnetic field \(\vec{B}=-(0.120 \mathrm{~T}) \hat{k}\). At a particular instant of time the velocity of the particle is \(\vec{V}=\left(1.05 \times 10^{6} \mathrm{~m} / \mathrm{s}\right)(-3 \hat{\boldsymbol{i}}+4 \hat{\boldsymbol{j}}+12 \hat{\boldsymbol{k}})\) and the force \(\vec{F}\) on the particle has a magnitude of \(2.45 \mathrm{~N}\). (a) Determine the charge \(q .\) (b) Determine the acceleration \(\vec{a}\) of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature \(R\) of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the \(x\) - and \(y\)-coordinates do vary in a periodic way. If the coordinates of the particle at \(t=0\) are \((x, y, z)=(R, 0,0)\), determine its coordinates at a time \(t=2 T\), where \(T\) is the period of the motion in the \(x y\)-plane.

Step by step solution

01

Calculate Charge q

The magnetic force on a charged particle moving in a magnetic field is given by \( \vec{F} = q (\vec{V} \times \vec{B}) \). To find the charge \( q \), first calculate the cross product \( \vec{V} \times \vec{B} \). Given \( \vec{V} = (1.05 \times 10^6 \mathrm{~m/s})(-3 \hat{i} + 4 \hat{j} + 12 \hat{k}) \) and \( \vec{B} = -(0.120 \mathrm{~T}) \hat{k} \), \[ \vec{V} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1.05\times 10^6 \times (-3) & 1.05\times 10^6 \times 4 & 1.05\times 10^6 \times 12 \ 0 & 0 & -0.120 \end{vmatrix} \]Calculate the determinant to find the result of the cross product:\[ \vec{V} \times \vec{B} = \hat{i}(4 \cdot -0.120 - 12 \cdot 0) - \hat{j}((-3) \cdot -0.120) + \hat{k}(0) = -0.504 \hat{i} - 0.378 \hat{j} \]The force magnitude is given as \( 2.45 \mathrm{~N} \), so equate it to calculate \( q \):\[ 2.45 = |q| \cdot \sqrt{(-0.504)^2 + (-0.378)^2} \]After calculation, \( q \approx -5.0 \times 10^{-19} \mathrm{~C} \).

02

Calculate Acceleration a

The acceleration of the particle can be calculated using Newton's second law \( \vec{F} = m \vec{a} \), where \( m = 2.58 \times 10^{-15} \mathrm{~kg} \) and \( \vec{F} \) is the magnetic force calculated in the previous step.The magnetic force vector was determined as:\[ \vec{F} = q (\vec{V} \times \vec{B}) = q (-0.504 \hat{i} - 0.378 \hat{j}) \]Find the magnitude of \( \vec{F} \) (as calculated previously, \( 2.45 \mathrm{~N} \)) and use it to calculate acceleration:\[ 2.45 \mathrm{~N} = 2.58 \times 10^{-15} \mathrm{~kg} \cdot \vec{a} \]\[ \vec{a} = \frac{2.45 \mathrm{~N}}{2.58 \times 10^{-15} \mathrm{~kg}} \approx 9.50 \times 10^{14} \mathrm{~m/s^2} \]

03

Determine Helical Path and Radius of Curvature

The particle experiences force only perpendicular to its velocity, thus causing circular motion in the plane perpendicular to \( \vec{B} \). As \( \vec{B} \) is along the z-axis, circular motion occurs in the xy-plane and the z-component of its velocity contributes to the helical motion.The radius \( R \) of the circle can be found using\[ R = \frac{m |V_{\perp}|}{|q|B} \]where \( V_{\perp} = \sqrt{(-3 \times 1.05 \times 10^6)^2 + (4 \times 1.05 \times 10^6)^2} = 5.25 \times 10^6 \mathrm{~m/s} \).Substitute values:\[ R = \frac{2.58 \times 10^{-15} \mathrm{~kg} \cdot 5.25 \times 10^6 \mathrm{~m/s}}{5.0 \times 10^{-19} \mathrm{~C} \cdot 0.120 \mathrm{~T}} \approx 2.26 \times 10^{-2} \mathrm{~m} \]

04

Find Cyclotron Frequency

The cyclotron frequency \( f_c \) is given by\[ f_c = \frac{|q|B}{2\pi m} \]Using known values:\[ f_c = \frac{5.0 \times 10^{-19} \cdot 0.120}{2\pi \times 2.58 \times 10^{-15}} \approx 3.7 \times 10^4 \mathrm{~Hz} \]

05

Determine Coordinates at Time 2T

The period \( T \) for circular motion in the \( xy \)-plane is given by\[ T = \frac{1}{f_c} \approx \frac{1}{3.7 \times 10^4} \mathrm{~s} \approx 2.70 \times 10^{-5} \mathrm{~s} \]At \( t = 2T \), the x and y coordinates complete two full cycles and return to the starting point \( (R,0) \).For the z-coordinate, which progresses linearly, calculate\[ z(t) = V_z \cdot 2T = (12 \times 1.05 \times 10^6) \cdot 2 \times 2.70 \times 10^{-5} \approx 6.78 \times 10^{2} \mathrm{~m} \]Therefore, at \( t = 2T \), the coordinates are \((R, 0, 678 \mathrm{~m})\).

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Key Concepts

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

Helical Motion of Charged Particles

When a charged particle moves through a magnetic field, it tends to follow a helical path if its velocity has components both parallel and perpendicular to the magnetic field. This is due to the magnetic force acting perpendicular to the particle’s velocity. Thus, the charged particle spirals along the direction of the magnetic field while maintaining a constant component of velocity in that direction.

In this case, the velocity of the particle can be broken into two components: one in the xy-plane, which is perpendicular to the magnetic field, and the other along the z-axis, parallel to the magnetic field. The component in the xy-plane causes the particle to move in a circular path, while the parallel component allows it to move linearly along the direction of the magnetic field, creating a helical motion.

• The circular component of the motion is influenced by the magnetic force.
• The linear component depends solely on the initial velocity of the particle along the magnetic field line.

This combined movement produces a helix, and is a fundamental aspect of how charged particles behave in magnetic fields.

Lorentz Force

The Lorentz force is a crucial concept in understanding how charged particles interact with magnetic fields. It describes the force exerted on a charged particle moving through magnetic and electric fields, and is given by the formula:

\( \vec{F} = q(\vec{V} \times \vec{B}) \)

where \(q\) is the charge of the particle, \(\vec{V}\) is the velocity, and \(\vec{B}\) is the magnetic field. This cross product leads to a force that is always perpendicular to both the velocity of the particle and the magnetic field.

• Because the force is perpendicular, it does not do work on the particle, meaning it does not change the particle's speed, only its direction.
• This perpendicular force is what causes circular motion in a plane perpendicular to the magnetic field.

Understanding the Lorentz force is key to predicting and explaining the spiral paths of charged particles in magnetic fields, as it highlights why the motion is circular around the field lines, influencing the particle's trajectory significantly.

Cyclotron Frequency

The cyclotron frequency is the frequency at which a charged particle orbits in a circular path in a perpendicular magnetic field. It is defined by the equation:

\( f_c = \frac{|q|B}{2 \pi m} \)

where \(f_c\) is the cyclotron frequency, \(q\) is the charge, \(B\) is the magnetic field strength, and \(m\) is the mass of the particle. This frequency provides insight into the periodic nature of the circular motion component of helical motion.

• The cyclotron frequency is only dependent on the charge-to-mass ratio and the magnetic field strength, not the speed of the particle.
• This frequency allows for prediction of the time it takes for the particle to complete one loop or cycle in its orbit.

This frequency is fundamental in applications involving charged particles in magnetic fields, such as particle accelerators, helping to understand and calculate the particle's behavior.

Circular Motion in Magnetic Fields

Circular motion occurs when the magnetic force provides the centripetal force necessary to keep a charged particle in a circular path perpendicular to the magnetic field. When a charged particle enters a magnetic field that is directed along a certain axis, it would tend to move in circles on a plane perpendicular to this axis.

This circular motion can be calculated using the formula for the radius of curvature:

\( R = \frac{m |V_{\perp}|}{|q|B} \)

where \(R\) is the radius of the circle, \(m\) is the mass, \(|V_{\perp}|\) is the component of velocity perpendicular to the magnetic field, \(q\) is the charge, and \(B\) is the magnetic field strength. The centripetal nature of the magnetic force allows the particle to stay in a circular path, with:

• The magnetic force constantly changing the direction of the velocity without affecting its magnitude.
• This component of motion occurring alongside linear motion along the magnetic field to create the helical path.

Understanding this circular motion is crucial for grasping the full motion of charged particles in a magnetic field.