Problem 21 - University Physics with Modern Physics

Question

A deuteron (the nucleus of an isotope of bydrogen) 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 to potential difference would the deuteron have to be accelerated to acquire this speed?

Step-by-Step Solution

Answer

Speed: \( 3.32 \times 10^6 \) m/s, Time for half revolution: \( 6.58 \times 10^{-9} \) s, Potential difference: \( 114.5 \) V.

Step 1: Understand the centripetal force condition

In a magnetic field, the force on a charged particle is given by \( F = qvB \), where \( q \) is the charge, \( v \) is the speed, and \( B \) is the magnetic field. This force provides the centripetal force for the circular motion, so \( qvB = \frac{mv^2}{r} \), where \( m \) is the mass and \( r \) is the radius of the path.

Step 2: Isolating for Speed

We need to find \( v \) (the speed of the deuteron). This can be done from the equation \( qvB = \frac{mv^2}{r} \). Rearranging, we find \( v = \frac{qBr}{m} \). The charge \( q \) of a deuteron is equal to the elementary charge \( e = 1.6 \times 10^{-19} \) C.

Step 3: Calculate the Speed

Substitute the given values into the equation: \[v = \frac{(1.6 \times 10^{-19} \, \mathrm{C})(2.50 \, \mathrm{T})(6.96 \times 10^{-3} \, \mathrm{m})}{3.34 \times 10^{-27} \, \mathrm{kg}} \approx 3.32 \times 10^6 \, \mathrm{m/s}\]Thus, the speed of the deuteron is approximately \( 3.32 \times 10^6 \, \mathrm{m/s} \).

Step 4: Determine the Time for Half a Revolution

The time for a complete revolution is \( T = \frac{2\pi r}{v} \). For half a revolution, the time \( t = \frac{\pi r}{v} \).

Step 5: Calculate the Time for Half a Revolution

Using \( r = 6.96 \times 10^{-3} \, \mathrm{m} \) and \( v = 3.32 \times 10^6 \, \mathrm{m/s} \), we have:\[t = \frac{\pi (6.96 \times 10^{-3} \, \mathrm{m})}{3.32 \times 10^6 \, \mathrm{m/s}} \approx 6.58 \times 10^{-9} \, \mathrm{s}\]It takes approximately \( 6.58 \times 10^{-9} \) s for half a revolution.

Step 6: Calculate the Potential Difference for Acceleration

The kinetic energy acquired is \( K = \frac{1}{2}mv^2 \). The work done by the electric potential is \( qV \), where \( V \) is the potential difference. Setting these equal, \( \frac{1}{2}mv^2 = qV \), solving for \( V \):\[ V = \frac{mv^2}{2q} \]

Step 7: Substitute Values to Find Potential Difference

Substitute in the given values: \[V = \frac{(3.34 \times 10^{-27} \, \mathrm{kg})(3.32 \times 10^6 \, \mathrm{m/s})^2}{2 \times 1.6 \times 10^{-19} \, \mathrm{C}} \approx 114.5 \, \mathrm{V}\]The potential difference required is approximately \( 114.5 \, \mathrm{V} \).

Key Concepts

Centripetal Force

When a deuteron moves in a magnetic field and follows a circular path, it experiences a force known as centripetal force. This force is essential to keep the deuteron moving in its circular path. In a magnetic field, any charged particle like a deuteron experiences a force given by the formula:

• \( F = qvB \)

Here, \( q \) is the charge, \( v \) is the speed of the particle, and \( B \) is the magnetic field strength.
The centripetal force is vital because it acts towards the center of the circular path, preventing the particle from flying off in a straight line. The balance between this magnetic force and the centripetal force required for circular motion is described by the equation:

• \( qvB = \frac{mv^2}{r} \)

Where \( m \) is the mass of the deuteron and \( r \) is the radius of the circular path. This relationship allows us to solve for the speed \( v \), which can be crucial for understanding the motion of particles in magnetic fields.

Magnetic Field

A magnetic field is an invisible area around a magnetic force where magnetic forces can be detected. It affects particles that have a charge, such as the deuteron in our example.

• When these charged particles move through a magnetic field, they experience a force that is perpendicular to both their direction of motion and the magnetic field lines.
• This perpendicular force causes the particles to execute circular or spiral paths.

The magnitude of this force can be calculated using the formula:
\( F = qvB \), where \( q \) is the charge, \( v \) is the speed of the particle, and \( B \) is the strength of the magnetic field.
This force is responsible for the circular motion of the deuteron and is always directed towards the center of the circle, providing the necessary centripetal force for its path.

Potential Difference

In physics, a potential difference, often called voltage, is what causes charged particles like deuterons to move. When a particle is accelerated by a potential difference, it gains kinetic energy. This relationship can be described by looking at work done by an electric field:

• The work done on a particle by an electric potential is given by \( qV \), where \( q \) is the charge and \( V \) is the potential difference.
• The kinetic energy gained is \( \frac{1}{2}mv^2 \).

By setting the work done equal to the kinetic energy gained, \( qV = \frac{1}{2}mv^2 \), we can solve for \( V \), the potential difference needed to reach a particular speed.
This is critical in applications like particle accelerators, where precise control of particle speed requires careful calculation of potential differences.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. For a moving deuteron in a magnetic field, this energy is crucial to understanding how fast the particle is moving. The formula to calculate kinetic energy is:

• \( K = \frac{1}{2}mv^2 \)

Where \( m \) represents the mass and \( v \) the speed of the deuteron.
As the deuteron moves faster, its kinetic energy increases. This relationship is key in determining how much potential difference is needed to accelerate the deuteron to a given speed.
Understanding kinetic energy also helps in studying energy conservation principles in systems where particles are continually gaining or losing energy as they interact with forces like magnetic fields.