Problem 19 - University Physics with Modern Physics

Question

A particle with charge \(6.40 \times 10^{-19} \mathrm{C}\) travels in a circular orbit with radius 4.68 \(\mathrm{mm}\) due to the force exerted on it by a magnetic field with magnitude 1.65 \(\mathrm{T}\) and perpendicular to the orbit. (a) What is the magnitude of the linear momentum \(\vec{p}\) of the particle? (b) What is the magnitude of the angular momentum \(\vec{L}\) of the particle?

Step-by-Step Solution

Answer

Linear momentum is approximately \(4.95 \times 10^{-21} \, \mathrm{kg\cdot m/s}\) and angular momentum is approximately \(2.32 \times 10^{-23} \, \mathrm{kg\cdot m^2/s}\).

Step 1: Understanding the Problem

The particle is moving in a circular path due to a magnetic field. We need to find its linear and angular momentum given its charge, orbit radius, and magnetic field strength.

Step 2: Applying the Lorentz Force Formula

The magnetic force acting on the charged particle in a magnetic field causes it to move in a circular path. The force can be given as: \[ F = qvB \]Where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field.

Step 3: Relating Force to Circular Motion

Since the particle moves in a circle, the magnetic force is also the centripetal force, given by: \[ F = \frac{mv^2}{r} \]Equating the two expressions for force:\[ qvB = \frac{mv^2}{r} \]Solve for \( v \), the velocity of the particle:\[ v = \frac{qBr}{m} \]

Step 4: Finding Linear Momentum

The linear momentum \( \vec{p} \) is given by the equation:\[ \vec{p} = mv \]Using the expression for velocity from the previous step, substituting \( v = \frac{qBr}{m} \):\[ \vec{p} = m \left( \frac{qBr}{m} \right) = qBr \]So, the magnitude of the linear momentum is \( p = qBr \).

Step 5: Plug in Values for Linear Momentum

Substitute the given values into \( p = qBr \):\[ q = 6.40 \times 10^{-19} \, \mathrm{C}, \]\[ B = 1.65 \, \mathrm{T}, \]\[ r = 4.68 \times 10^{-3} \, \mathrm{m} \]Calculate:\[ p = (6.40 \times 10^{-19})(1.65)(4.68 \times 10^{-3}) \, \mathrm{kg \cdot m/s} \]

Step 6: Calculation of Linear Momentum

Compute the value of the linear momentum:\[ p \approx 4.95 \times 10^{-21} \, \mathrm{kg \cdot m/s} \]

Step 7: Applying the Angular Momentum Formula

Angular momentum \( \vec{L} \) for an object moving in a circle is given by:\[ \vec{L} = r \times \vec{p} \]Where \( r \) is the radius and \( \vec{p} \) is the linear momentum.

Step 8: Substitute Values for Angular Momentum

Using the magnitude of \( \vec{p} \) calculated:\[ L = rp = (4.68 \times 10^{-3} \, \mathrm{m})(4.95 \times 10^{-21} \, \mathrm{kg \cdot m/s}) \]

Step 9: Calculation of Angular Momentum

Compute the magnitude of the angular momentum:\[ L \approx 2.32 \times 10^{-23} \, \mathrm{kg \cdot m^2/s} \]

Key Concepts

Linear Momentum

Linear momentum is a fundamental concept in physics that measures the motion of an object. It is represented by the symbol \( \vec{p} \) and is calculated as the product of an object's mass \( m \) and its velocity \( v \). Here’s the simple formula:

• \( \vec{p} = mv \)

In the context of charged particles moving in a magnetic field, the velocity is determined by the circular path the particle takes. This is due to the magnetic force acting as the centripetal force.
By substituting the expression for velocity we have, the linear momentum becomes \( p = qBr \), where \( q \) is the charge, \( B \) is the magnetic field strength, and \( r \) is the radius of the circular path.
Thus, it encapsulates how quickly and in what direction the particle is traveling.

Angular Momentum

Angular momentum \( \vec{L} \) quantifies the extent of rotation a body exhibits around a point or axis. For objects moving in a circular path, angular momentum arises from both the mass of the object and its velocity. The formula is:

• \( \vec{L} = r \times \vec{p} \)

Where \( r \) is the radius of the circular path, and \( \vec{p} \) is the linear momentum. In this context, the angular momentum of a charged particle in a magnetic field can be expressed as,

• \( L = rp = r(qBr) \)

Here, the emphasis is on how the particle's linear momentum interacts with the circle it travels, creating a rotational motion considered in angular momentum.

Lorentz Force

The Lorentz force is the fundamental force experienced by a charged particle in a magnetic field, described by the formula:

• \( F = qvB \)

Where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field. This force is always perpendicular to both the velocity of the particle and the direction of the magnetic field, causing the particle to move in a circular path.
By balancing this with the centripetal force required to maintain circular motion \( F = \frac{mv^2}{r} \), we derive the particle’s velocity and, subsequently, its linear momentum via \( v = \frac{qBr}{m} \). This interplay brings insights into how magnetic fields can control charged particles in devices and applications.

Circular Motion

Circular motion occurs when an object moves along a circular path. In the presence of a magnetic field, a charged particle executes circular motion due to the radial magnetic force. This force ensures that the particle remains in its circular trajectory by providing the centripetal force required.
Crucially, its velocity and the radius of the path play roles in determining the magnitude of centripetal force needed, given by \( F = \frac{mv^2}{r} \). This relationship ties directly into linear momentum, as the force imparted by the magnetic field transforms into the centripetal force maintaining the motion.
Consequently, understanding circular motion helps reveal how charged particles can be controlled in scientific experiments and technical applications, like cyclotrons and mass spectrometers.