Problem 21 - University Physics with Modern Physics

Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) and a charge of \(+e .\) The deuteron travels in a circular path with a radius of 6.96 \(\mathrm{mm}\) in a magnetic field with magnitude 2.50 \(\mathrm{T}\) (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Step-by-Step Solution

Answer

(a) The speed is \( 8.34 \times 10^5 \, m/s \). (b) Time for half a revolution is \( 2.62 \times 10^{-8} \, s \). (c) Potential difference is \( 1457 \, V \).

Step 1: Understanding the Circular Motion

In a magnetic field, a charged particle moving with velocity \( v \) experiences a force, called the Lorentz force, which causes it to move in a circular path. For a particle of charge \( q \), mass \( m \), and speed \( v \) moving in a magnetic field \( B \), the force is \( F = qvB \). For circular motion, this force is the centripetal force, \( F_c = \frac{mv^2}{r} \). Equating these gives us the equation for speed: \[ qvB = \frac{mv^2}{r}. \]

Step 2: Solve for Speed

Rearrange the formula \( qvB = \frac{mv^2}{r} \) to solve for \( v \): \[ v = \frac{qBr}{m}. \] Plug in the known values: \( q = 1.6 \times 10^{-19} \, C \), \( B = 2.50 \, T \), \( r = 6.96 \times 10^{-3} \, m \), and \( m = 3.34 \times 10^{-27} \, kg \). Compute the value of \( v \): \[ v = \frac{(1.6 \times 10^{-19})(2.50)(6.96 \times 10^{-3})}{3.34 \times 10^{-27}} = 8.34 \times 10^5 \, m/s. \]

Step 3: Calculate Time for Half a Revolution

The time for one full revolution is the circumference of the circle divided by the speed, \( T = \frac{2\pi r}{v} \). For half a revolution, the time is half of this: \( t = \frac{\pi r}{v} \). Substitute \( r = 6.96 \times 10^{-3} \, m \) and \( v = 8.34 \times 10^5 \, m/s \) to get: \[ t = \frac{\pi (6.96 \times 10^{-3})}{8.34 \times 10^5} = 2.62 \times 10^{-8} \, s. \]

Step 4: Potential Difference Calculation

The electric potential energy given by a potential difference \( V \) is equated to the kinetic energy, \( KE = \frac{1}{2}mv^2 \), as \( qV = \frac{1}{2}mv^2 \). Solve for \( V \): \[ V = \frac{mv^2}{2q}. \] Substitute the known values \( m = 3.34 \times 10^{-27} \, kg \), \( v = 8.34 \times 10^5 \, m/s \), and \( q = 1.6 \times 10^{-19} \, C \): \[ V = \frac{(3.34 \times 10^{-27})(8.34 \times 10^5)^2}{2(1.6 \times 10^{-19})} = 1.457 \times 10^3 \, V. \]

Key Concepts

Circular Motion

Circular motion is the movement of a particle along the circumference of a circle. When a charged particle, like a deuteron, moves through a magnetic field, it does so in a circular manner. This is due to the magnetic force acting perpendicular to the velocity, constantly changing the particle's direction. While the speed remains constant in uniform circular motion, the direction keeps changing, forming a circle. In this scenario, the magnetic field provides a centripetal force that maintains the particle's trajectory. Understanding circular motion is key to predicting how particles behave in magnetic fields.

Lorentz Force

The Lorentz force is a fundamental concept when dealing with charged particles in a magnetic field. It is the force experienced by a charged particle due to electromagnetic fields. For a particle with charge \( q \), moving with velocity \( v \) in a magnetic field \( B \), the force is given by the equation:

• \( F = qvB \)

This force acts perpendicular to both the magnetic field and the velocity of the particle, influencing the path without altering the speed. In the context of a deuteron in a magnetic field, the Lorentz force causes the deuteron to exhibit circular motion. It is crucial for determining the radius of the circular path and understanding the nature of particle trajectories in magnetic fields.

Potential Difference

Potential difference, or voltage, is a measure of the work done to move a charge from one point to another in an electric field. In magnetic applications, potential difference is often related to kinetic energy conversion. A deuteron moving through a potential difference gains kinetic energy corresponding to that difference. The relation between potential difference \( V \), charge \( q \), and kinetic energy \( KE \) is expressed as:

• \( qV = \frac{1}{2}mv^2 \)

By rearranging, we find the potential difference required for a deuteron to reach a specific speed. This is essential in experiments where particles need to reach specific velocities for reactions or observations.

Centripetal Force

Centripetal force is the force that keeps an object moving in a circular path. In magnetic applications, it is provided by the magnetic force itself. For a particle in circular motion, the centripetal force

• \( F_c = \frac{mv^2}{r} \)

This force must balance with the magnetic force for the particle to maintain its circular path. The balance of forces for a charged particle in a magnetic field can be summarized as:

• \( qvB = \frac{mv^2}{r} \)

Understanding this balance helps in determining the speed and radius of the trajectory, crucial for applications in cyclotrons and other particle accelerators.

Kinetic Energy

Kinetic energy (KE) represents the energy possessed by an object due to its motion and is expressed by the formula:

• \( KE = \frac{1}{2}mv^2 \)

For a deuteron moving in a magnetic field, its kinetic energy is the result of acceleration through a potential difference. The kinetic energy can be used to determine the speed of the deuteron as it translates energy from the electric field into motion. Analyzing kinetic energy is essential for understanding how particles like deuterons can be manipulated and accelerated within fields for research and technological applications. The conversion of energy types is pivotal in many experimental settings, leading to discoveries about particle behavior at different velocities.