Problem 61 - University Physics with Modern Physics

Question

A particle with negative charge \(q\) and mass \(m=2.58 \times\) 10 \(^{-15} \mathrm{kg}\) is traveling through a region containing a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(0.120 \mathrm{T}) \hat{\boldsymbol{k}} .\) At a particular instant of time the velocity of the particle is \(\overrightarrow{\boldsymbol{v}}=\left(1.05 \times 10^{6} \mathrm{m} / \mathrm{s}\right)(-3 \hat{\imath}+4 \hat{\jmath}+\) 12\(\hat{k} )\) and the force \(\vec{F}\) on the particle has a magnitude of 1.25 \(\mathrm{N}\) . (a) Determine the charge \(q\) . (b) Determine the acceleration \(\overrightarrow{\boldsymbol{d}}\) 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

Answer

Charge: \(1.44 \times 10^{-5} \text{ C}\); Radius: \(6 \times 10^{-3} \text{ m}\); Cyclotron frequency: \(1.11 \times 10^9 \text{ Hz}\).

Step 1: Calculate the Magnetic Force using Cross Product

The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force formula: \( \vec{F} = q(\vec{v} \times \vec{B}) \). To find the charge, we must first calculate the cross product \( \vec{v} \times \vec{B} \). Given \( \vec{v} = (1.05 \times 10^6 \text{ m/s})(-3\hat{i} + 4\hat{j} + 12\hat{k}) \) and \( \vec{B} = -(0.120 \text{ T}) \hat{k} \), the cross product found using the determinant is \( \vec{v} \times \vec{B} = 1.05 \times 10^6 (-3 \cdot 0.12\hat{j} + 4 \cdot 0.12\hat{i}) \), which simplifies to \(( -0.50 \hat{j} + 0.72 \hat{i}) \times 10^5 \text{ T m/s} \).

Step 2: Calculate the Magnitude of the Force

The force magnitude given is 1.25 N, and we have \( \vec{F} = q\vec{v} \times \vec{B} \). Calculate the magnitude: \( |\vec{v} \times \vec{B}| = \sqrt{(0.72 \times 10^5)^2 + (-0.50 \times 10^5)^2} = 0.87 \times 10^5 \text{ T m/s} \). Use \( F = |q||\vec{v} \times \vec{B}| \) to find \( q = \frac{F}{|\vec{v} \times \vec{B}|} = \frac{1.25}{0.87 \times 10^5} \approx 1.44 \times 10^{-5} \text{ C} \).

Step 3: Find Acceleration Using Newton's Second Law

Acceleration \( \vec{a} \) can be found using \( \vec{F} = m\vec{a} \). With \( m = 2.58 \times 10^{-15} \text{ kg} \) and \( |\vec{F}| = 1.25 \text{ N} \), rearrange to find \( \vec{a} = \frac{\vec{F}}{m} = \frac{1.25}{2.58 \times 10^{-15}} \approx 4.84 \times 10^{14} \text{ m/s}^2 \).

Step 4: Describe Motion Path and Calculate Radius of Curvature

The force is perpendicular to velocity, causing circular motion in the \(xy\)-plane, creating a helical path. The radius \( R \) of the circle is found using \( R = \frac{mv_\perp}{|q|B} \), where \( v_\perp = |\vec{v}\text{ in }xy\)| is \((3^2 + 4^2)^{0.5}\cdot 1.05 \times 10^6 \). Calculate \( R \approx \frac{2.58 \times 10^{-15}\cdot 5.0 \times 10^6}{1.44 \times 10^{-5} \cdot 0.12} \approx 6 \times 10^{-3} \text{ m}\).

Step 5: Calculate Cyclotron Frequency

Cyclotron frequency \( f_c \) is given by \( f_c = \frac{|q|B}{2\pi m} = \frac{1.44 \times 10^{-5} \cdot 0.12}{2\pi \times 2.58 \times 10^{-15}} \approx 1.11 \times 10^9 \text{ Hz} \).

Step 6: Determine Particle Position at Time \( 2T \)

The period \( T \) in the \(xy\)-plane is \( T = \frac{1}{f_c} = \frac{1}{1.11 \times 10^9} \approx 9.0 \times 10^{-10} \text{ s} \). For full periods \( 2T \), \( xy\)-motion returns to initial, so \( z \) moves uniformly with speed \( v_z \cdot 2T \), where \( v_z = 12 \times 1.05 = 1.26 \times 10^7 \text{ m/s} \). Therefore, \( z = v_z \cdot 2T \).

Key Concepts

Cyclotron Frequency

The cyclotron frequency is a crucial concept when examining the motion of charged particles in a magnetic field. It describes how fast a charged particle, like an electron or proton, orbits in a perpendicular plane under the influence of a magnetic field. This frequency is independent of speed and only depends on the charge-to-mass ratio and the magnetic field's strength.

This frequency can be mathematically expressed as:\[ f_c = \frac{|q|B}{2\pi m} \]
where:

• \( f_c \) is the cyclotron frequency,
• \( |q| \) is the absolute value of the charge,
• \( B \) is the magnetic field strength,
• \( m \) is the mass of the particle.

In the case of the negative particle described in the exercise, substituting the given values into the formula produces a frequency of approximately \( 1.11 \times 10^9 \text{ Hz} \). This incredible frequency indicates how rapidly the particle orbits in the magnetic field. Understanding this helps visualize and predict the particle's behavior under magnetic influences.

Helical Motion

Helical motion arises when a charged particle moves through a uniform magnetic field with a velocity that has both perpendicular and parallel components to the field. This movement results in a spiral path, mixing circular and linear motion seen often in magnetic fields.

In simple terms, when the velocity of a particle is not completely perpendicular to the magnetic field, the particle starts spiraling around the magnetic field lines. This spiraling is called helical motion and is observable due to the Lorentz force affecting only the perpendicular velocity component.

For example, when the exercise's charged particle moves through the magnetic field, it follows a helical path, as it has components of velocity both in the plane perpendicular to the magnetic field and along the direction of the field. This complex motion illustrates why particles in magnetic fields generally do not move linearly.

Radius of Curvature

The radius of curvature is another essential concept that describes the circular aspect of the helical motion of charged particles in a magnetic field. It specifically refers to how tightly or loosely a particle moves in its circular path perpendicular to the magnetic field lines.

The radius of curvature can be calculated using the formula:\[ R = \frac{mv_\perp}{|q|B} \]
where:

• \( R \) is the radius of curvature,
• \( m \) is the particle's mass,
• \( v_\perp \) is the component of the velocity perpendicular to the magnetic field,
• \( |q| \) is the absolute charge,
• \( B \) represents the magnetic field's strength.

In the given exercise, this radius is found to be approximately \( 6 \times 10^{-3} \text{ m} \), indicating the size of the spirals in the helical path the charged particle takes. The radius of curvature helps understand how magnetic fields confine particle motion and influence the geometric nature of trajectories.

Magnetic Fields

A magnetic field, a fundamental concept in physics, represents a region around a magnetic material or a moving electric charge within which the force of magnetism acts. These fields are crucial in determining the behavior of charged particles and dictating how they move.

Magnetic fields are characterized by their strength and direction. In this exercise, the magnetic field is uniform and directed in the negative z-direction, measured as \(B = 0.120 \text{ T} \).

When a charged particle moves through a magnetic field, it experiences a force described by the Lorentz force: \( \vec{F} = q(\vec{v} \times \vec{B}) \). This force is always perpendicular to both the velocity of the particle and the magnetic field, resulting in circular or helical motion rather than straight-line paths.

Understanding magnetic fields helps predict and analyze the motion of charged particles, enabling applications in areas like cyclotrons, magnetic confinement, and even auroras seen in Earth's atmosphere.