Problem 29 - University Physics with Modern Physics

Question

A \(150-\mathrm{V}\) battery is connected across two parallel metal plates of area 28.5 \(\mathrm{cm}^{2}\) and separation \(8.20 \mathrm{mm} .\) A beam of alpha particles (charge \(+2 e,\) mass \(6.64 \times 10^{-27} \mathrm{kg} )\) is accelerated from rest through a potential difference of 1.75 \(\mathrm{kV}\) and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

Step-by-Step Solution

Answer

The magnetic field needed is approximately 0.109 T, directed into the plane of the page.

Step 1: Determine the Electric Field between the Plates

The electric field \( E \) between the plates is determined by the voltage \( V \) across the plates and the distance \( d \) between them. The formula is \[E = \frac{V}{d}\]Substitute \( V = 150\, \text{V} \) and \( d = 8.20\, \text{mm} = 8.20 \times 10^{-3} \text{m} \) to find \[E = \frac{150}{8.20 \times 10^{-3}} = 18292.68 \text{ V/m}.\]

Step 2: Calculate the Velocity of Alpha Particles

The velocity \( v \) of the alpha particles after being accelerated through a potential difference \( V_a = 1750 \text{ V} \) can be found using the energy equation \[qV_a = \frac{1}{2}mv^2\]where \( q = 2e = 2(1.6 \times 10^{-19} \text{ C}) \) and \( m = 6.64 \times 10^{-27} \text{ kg} \). Solve for \( v \) as follows: \[2e \cdot 1750 = \frac{1}{2} \cdot 6.64 \times 10^{-27} \cdot v^2\]\[ v = \sqrt{\frac{2 \cdot 2 \cdot 1.6 \times 10^{-19} \cdot 1750}{6.64 \times 10^{-27}}} = 1.68 \times 10^5 \text{ m/s}.\]

Step 3: Set the Force Equations for No Deflection

For the alpha particles to be undeflected, the magnetic force \( F_B \) must equal the electric force \( F_E \). Therefore, \( F_E = E \cdot q \) and \( F_B = q \cdot v \cdot B \). Set these equal to each other:\[ E \cdot q = q \cdot v \cdot B \]Simplifying gives:\[ B = \frac{E}{v}\]Substitute \( E = 18292.68 \text{ V/m} \) and \( v = 1.68 \times 10^5 \text{ m/s} \).

Step 4: Calculate the Magnetic Field Magnitude

Plug in the values for \( E \) and \( v \) from the previous steps to find the magnetic field \( B \):\[B = \frac{18292.68}{1.68 \times 10^5} \approx 0.109 \text{ T}.\]

Step 5: Determine the Direction of Magnetic Field

Using the right-hand rule, with the velocity of positive charges to the right and electric field upward (pointing from positive to negative plate), the magnetic field required to produce zero net force must be directed into the plane of the page.

Key Concepts

Electric Field

In physics, the electric field is a crucial concept that describes the region around a charged object where other charges would feel a force. It's a vector quantity, meaning it has both magnitude and direction. The electric field between two parallel plates, like those in our exercise, is uniform and can be calculated easily using the formula:
\[E = \frac{V}{d}\]This formula shows that the electric field strength \( E \) is directly proportional to the voltage \( V \) across the plates and inversely proportional to the separation \( d \) between them.

• Voltage across the plates: 150 V
• Separation between the plates: 8.20 mm or 8.20 x 10^-3 m

This setup creates an electric field that affects any charged particles entering the region. The electric field's direction is typically from the positive to the negative plate, which means any positive charge will naturally try to move in the direction of the field.

Alpha Particles

Alpha particles are a type of nuclear radiation consisting of two protons and two neutrons. This gives them a charge of \(+2e\) (where \( e \) represents the elementary charge of approximately \(1.6 \times 10^{-19}\) C) and a relatively large mass for a particle (6.64 \times 10^{-27} \mathrm{kg}).

• Charge: \(+2e\)
• Mass: \(6.64 \times 10^{-27} \mathrm{kg}\)

Alpha particles do not penetrate deeply into materials, but they can cause significant damage at the atomic level where they do impact.
In the context of the exercise, these alpha particles are first accelerated from rest through a potential difference, gaining kinetic energy in the process.

Potential Difference

A potential difference, often called voltage, is the difference in electrical potential energy between two points in a circuit. It's the driving factor behind the movement of charge carriers in an electric field. In our context, a potential difference of 1.75 kV accelerates the alpha particles.

• Potential Difference (acceleration): 1.75 kV

When particles are subjected to a potential difference, their kinetic energy changes according to the equation:
\[qV = \frac{1}{2}mv^2\]For alpha particles, this equation tells us how much energy they've gained and thus, how fast they are moving after being accelerated. This change in speed is critical for determining the forces acting on them in the presence of electric and magnetic fields, as seen in the exercise.

Undeflected Path of Charged Particles

In the presence of both electric and magnetic fields, it's possible for charged particles to travel in a straight, undeflected path if the forces exerted by each field are equal in magnitude and opposite in direction. This setup requires that:
\[F_E = F_B\]Where \( F_E = E \times q \) is the electric force and \( F_B = q \times v \times B \) is the magnetic force.
For the problem at hand, equating these forces and solving for the magnetic field \( B \) yields:\[B = \frac{E}{v}\]This solution shows that the magnetic field strength depends on the speed of the particles and the electric field strength. Substituting the known values from the exercise provides the magnetic field magnitude. Moreover, the direction of the magnetic field is determined using the right-hand rule. When facing an "out of the page" force requirement, the magnetic field must be directed into the page to achieve zero net force on the alpha particles, allowing for an undeflected path.