Problem 63 - 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 \(\vec { \boldsymbol { B } } = - ( 0.120 \mathrm { T } ) \hat { \boldsymbol { k } }\) . At a particular instant of time the velocity of the particle is \(\vec { \boldsymbol { v } } = \left( 1.05 \times 10 ^ { 6 } \mathrm { m } / \mathrm { s } \right) ( - 3 \hat { \imath } + 4 \hat { \jmath } + 12 \hat { \boldsymbol { k } } )\) and the force \(\vec { \boldsymbol { F } }\) 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. (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

\( q = -4.08 \text{ C} \); \( \vec{a} = 9.5 \times 10^{14} \text{ m/s}^2 \); motion is helical because of parallel velocity to \( \vec{B} \); \( R \approx 2.74 \times 10^{-4} \text{ m} \); \( \omega_c \approx 1.89 \times 10^{13} \text{ rad/s} \); coordinates at \(t = 2T\) are \((R, 0, 8.4 \times 10^{-6})\).

Step 1: Calculate Lorentz Force Components

The Lorentz force acting on a charged particle moving in a magnetic field is given by the equation \( \vec{F} = q(\vec{v} \times \vec{B}) \). Given: \( \vec{B} = -0.120 \hat{k} \), \( \vec{v} = (1.05 \times 10^6)(-3 \hat{i} + 4 \hat{j} + 12 \hat{k}) \). First, calculate \( \vec{v} \times \vec{B} \).Compute \( \vec{v} \times \vec{B} \) using the determinant method:\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 4 & 12 \ 0 & 0 & -0.120 \end{vmatrix} = \hat{i}(4 \times -0.120) - \hat{j}(-3 \times -0.120) + \hat{k}(0) \]This simplifies to:\[ \vec{v} \times \vec{B} = -0.48 \hat{i} - 0.36 \hat{j} \] (in T·m/s)

Step 2: Solve for Charge q

The magnitude of the force is given as 2.45 N. Substituting into the Lorentz force equation: \[ 2.45 = q \sqrt{(-0.48)^2 + (-0.36)^2} \]Solving for \( q \):\[ 2.45 = q \times 0.6 \]\[ q = \frac{2.45}{0.6} = 4.0833 \text{ C} \] since the charge must be negative.

Step 3: Calculate Acceleration

Acceleration \( \vec{a} \) is given by \[ \vec{a} = \frac{\vec{F}}{m} \]. The magnitude of \( \vec{F} \) is 2.45 N and the mass \( m = 2.58 \times 10^{-15} \text{ kg} \).\[ \vec{a} = \frac{2.45}{2.58 \times 10^{-15}} = 9.4961 \times 10^{14} \text{ m/s}^2 \]

Step 4: Explain the Helical Path and Calculate Radius

The path is helical because the velocity has a component parallel to the magnetic field (\(12\hat{k}\)) and components perpendicular to it, causing circular motion in the plane \(xy\). Radius \( R \) of the circular motion is calculated by balance of centripetal force and Lorentz force:\[ |\vec{F}| = \frac{mv^2}{R} \text{ for perpendicular components} \]\[ R = \frac{mv_{\perp}}{qB} \text{, where } v_{\perp} = \sqrt{(-3)^2 + 4^2} \times 1.05 \times 10^6 \text{ m/s} \]\[ v_{\perp} = 5 \times 1.05 \times 10^6 \text{ m/s } = 5.25 \times 10^6 \text{ m/s} \]\[ R = \frac{(2.58 \times 10^{-15})(5.25 \times 10^6)}{4.0833 \times 0.120} \approx 2.74 \times 10^{-4} \text{ m} \]

Step 5: Determine Cyclotron Frequency

Cyclotron frequency \( \omega_c \) equals \( \frac{qB}{m} \).\[ \omega_c = \frac{4.0833 \times 0.120}{2.58 \times 10^{-15}} \approx 1.89 \times 10^{13} \text{ rad/s} \]

Step 6: Calculate Periodic Motion Coordinates at t = 2T

The period \( T \) in the \(xy\)-plane is \( \frac{2\pi}{\omega_c} \):\[ T = \frac{2\pi}{1.89 \times 10^{13}} \approx 3.33 \times 10^{-13} \text{ s} \]At \( t = 2T \), the positions in the \(xy\)-plane will have completed two cycles, thus returning to initial \(x, y\) positions: \( (R, 0) \), and moving along \(z\) by \( v_{\parallel} \times 2T \):\[ z = 12 \times 1.05 \times 10^6 \times 2 \times 3.33 \times 10^{-13} = 8.4 \times 10^{-6} \text{ m} \]Final coordinates are \( (R, 0, 8.4 \times 10^{-6}) \).

Key Concepts

Lorentz Force

The Lorentz force is a fundamental concept in electromagnetism, describing how charged particles move under the influence of electric and magnetic fields. Specifically, the force acting on a charged particle in a magnetic field can be expressed with the formula \( \vec{F} = q ( \vec{v} \times \vec{B} ) \), where \( q \) is the charge, \( \vec{v} \) is the velocity, and \( \vec{B} \) is the magnetic field.
This force is perpendicular to both the velocity of the particle and the magnetic field, ensuring that the speed of the particle remains constant if it's only subject to the magnetic field. Unlike an electric force, which can increase or decrease the speed of a particle, the magnetic component only alters its direction. Think of this as the force that curves the path of the particle, but doesn't let it speed up or slow down.
This principle is crucial in machines like cyclotrons and mass spectrometers, where manipulating charged particles is necessary for their operation. It helps us find the charge of a particle if the force and magnetic field are known.

Helical Motion

When a charged particle, such as our particle with velocity components \( -3 \hat{i} + 4 \hat{j} + 12 \hat{k} \), moves through a magnetic field, it experiences helical motion. This path results when there's a velocity component parallel to the magnetic field. In this case, the \( 12 \hat{k} \) component is parallel to the magnetic field direction and doesn't change due to the Lorentz force.
The motion perpendicular to the field, due to components \( -3 \hat{i} \) and \( 4 \hat{j} \), results in circular motion. Combined with the constant velocity component parallel to the field, the path traced by the particle is a helix—imagine a circular spring stretched along a line.
This is like the particle corkscrewing through space. Helical motion occurs naturally in environments like the Earth's magnetic field, guiding particles from solar winds in fascinating spiral trajectories.

Cyclotron Frequency

Cyclotron frequency is the rate at which a charged particle orbits in a magnetic field. It's an essential concept in understanding particle accelerators. Mathematically, it's expressed as \( \omega_c = \frac{qB}{m} \). This formula shows how the frequency depends on the charge \( q \), the magnetic field strength \( B \), and the particle's mass \( m \).
The cyclotron frequency is independent of the velocity of the particle, meaning it's constant for a specific particle in a given magnetic field. This repetition rate is important for applications that require synchronized particle motion, like in a cyclotron—an apparatus used to accelerate charged particles to high energies.
Think of the cyclotron frequency as the beat or rhythm at which the particle dances in a magnetic field. This rhythm allows for precise timing controls in scientific and medical technology.

Radius of Curvature

The radius of curvature is a measure of how sharply a path bends. In the context of a charged particle moving through a magnetic field, it represents the radius of the particle's circular path within the helical motion.
To find it, we use the formula \( R = \frac{mv_{\perp}}{qB} \), where \( m \) is the mass, \( v_{\perp} \) is the perpendicular component of velocity, \( q \) is the charge, and \( B \) is the magnetic field strength. This calculation helps determine how tightly the particle is spiraling.
This concept is key in designing devices like magnetic separators, which require precise control over the trajectories of charged particles. By adjusting the radius of curvature, scientists and engineers can filter particles based on their mass and charge, selectively isolating desired species.