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\(\bullet\) 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?

Short Answer

Expert verified
A magnetic field of approximately 0.0976 T with direction out of the page is needed.

Step by step solution

01

Understand the Electric Field between Plates

The electric field \( E \) between the plates is uniform and can be calculated using the formula \( E = \frac{V}{d} \), where \( V = 150 \, \text{V} \) is the voltage across the plates and \( d = 8.20 \, \text{mm} = 8.20 \times 10^{-3} \, \text{m} \) is the separation between the plates.
02

Calculate the Electric Field

Substitute the values into the formula:\[E = \frac{150}{8.20 \times 10^{-3}} = 18293.9 \, \text{V/m}\]
03

Analyze Forces Acting on Alpha Particles

When the alpha particles enter the electric field between the plates, the electric force \( F_e \) acts on them, given by \( F_e = qE \), where \( q = 2e = 2 \times 1.60 \times 10^{-19} \, \text{C} \). To prevent deflection, the magnetic force \( F_m = qvB \) must balance the electric force, thus \( qE = qvB \).
04

Determine Velocity of Alpha Particles

The velocity \( v \) of the alpha particles can be determined from the kinetic energy gained by accelerating through a potential difference of 1.75 kV. The energy is \( qV = \frac{1}{2} mv^2 \), where \( m = 6.64 \times 10^{-27} \, \text{kg} \) is the mass of an alpha particle. Solve for \( v \).
05

Calculate Alpha Particles' Velocity

Substitute the values into the kinetic energy equation:\[qV = \frac{1}{2}mv^2 \implies 2e \times 1.75 \times 10^3 = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2\]Solving for \( v \):\[v = \sqrt{\frac{2 \times 2 \times 1.75 \times 10^3 \times 1.60 \times 10^{-19}}{6.64 \times 10^{-27}}}\]\[v \approx 1.87 \times 10^5 \text{ m/s}\]
06

Calculate Magnitude of Magnetic Field

Using the balance of forces \( qE = qvB \) and solving for \( B \), we find:\[B = \frac{E}{v} = \frac{18293.9}{1.87 \times 10^5} \approx 0.0976 \, \text{T}\]
07

Determine Direction of Magnetic Field

The direction of the magnetic field should be perpendicular to both the velocity of the alpha particles and the electric field to provide a magnetic force in the opposite direction to the electric force. Using the right-hand rule, the magnetic field should point into or out of the page depending on the conventional current direction. If particles have positive charge and positively charged plates placed on top, using right-hand rule, magnetic field points out of the page.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
When charged particles like alpha particles enter an electric field, they experience a force called the electric force. This force acts on the particles because of their charge and the presence of an electric field between two points. The electric force, denoted by \( F_e \), can be calculated using the formula: \[ F_e = qE \] where \( q \) is the charge of the particle and \( E \) is the strength of the electric field. For our exercise, alpha particles have a charge of \( +2e \), which means they experience double the basic charge of an electron.
  • The electric force direction depends on the sign of the charge and the field direction.
  • Positive charges like alpha particles are pushed in the direction of the field.
  • This force can accelerate or deflect the particles unless counterbalanced.
In the case of our problem, the electric force is balanced by the magnetic force, ensuring the alpha particles move straight as they pass between the plates.
Magnetic Force
Just as the electric field exerts a force on charged particles, a magnetic field also exerts a force known as the magnetic force. This force, \( F_m \), acts perpendicular to both the velocity \( v \) of the charged particle and the magnetic field \( B \). It can be calculated using the formula: \[ F_m = qvB \] This force ensures that the charged particles do not deviate from their path when entering a magnetic field.
  • The magnetic force direction can be predicted using the right-hand rule.
  • For a positive charge, point your thumb in the direction of velocity, fingers in the direction of the magnetic field; the palm then shows the force direction.
  • In our case, the magnetic field needed to be just the right strength and direction to counter the electric force.
By balancing these forces, the alpha particles remain undeflected between the plates.
Electric Field
An electric field describes the area around charged objects where other charges would experience an electric force. It is often generated between two parallel plates connected to a voltage difference. The electric field \( E \) is given by: \[ E = \frac{V}{d} \] where \( V \) is the voltage across the plates, and \( d \) is the distance separating them.
  • The electric field is uniform in strength and direction between parallel plates.
  • Its direction is defined as from the positive plate to the negative plate.
  • In our exercise, the electric field plays a crucial role in pushing the alpha particles.
Understanding the electric field helps determine the force exerted on the charge, facilitating calculations for other forces required for the equilibrium.
Alpha Particles
Alpha particles are positively charged particles composed of two protons and two neutrons, much like a helium nucleus. They have a charge of \( +2e \) and are relatively heavy compared to other subatomic particles. Their mass is \( 6.64 \times 10^{-27} \text{ kg} \). Due to their composition, they play a significant role in the study of both electric and magnetic fields.
  • Alpha particles, being positively charged, are influenced by both electric and magnetic fields.
  • In our problem, they enter the region between the plates perpendicularly to the electric field.
  • Their behavior in fields helps scientists understand and measure field strengths.
The dynamics of alpha particles show the principles of electromagnetic interactions, often used in fields like nuclear physics and engineering.
Kinetic Energy
Kinetic energy is the energy possessed by a particle due to its motion. When a charged particle accelerates through a potential difference, it gains kinetic energy, which can be calculated using: \[ KE = \frac{1}{2}mv^2 \] For charged particles, it is also given by the electrical work done on them: \[ KE = qV \] where \( V \) is the accelerating potential difference.
  • In the exercise, alpha particles are accelerated from rest, converting the electrical energy to kinetic energy.
  • Their velocity is derived from the gained kinetic energy, calculated from the potential difference.
  • This velocity is crucial for determining the magnetic field strength needed for stability between the plates.
Kinetic energy calculations reveal how fast a particle moves and how it can be influenced by external fields, serving as a fundamental aspect in mechanics and electromagnetism.

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Most popular questions from this chapter

\bullet A 150 g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a 125 vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal 0.250 T magnetic field directed from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

\(\bullet\) An electric bus operates by drawing current from two parallel overhead cables, at a potential difference of \(600 \mathrm{V},\) and spaced 55 \(\mathrm{cm}\) apart. When the power input to the bus's motor is at its maximum power of \(65 \mathrm{hp},(\) a) what current does it draw and (b) what is the attractive force per unit length between the cables?

\(\bullet\) Singly ionized (onne electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 \(\mathrm{V} / \mathrm{m}\) and the magnetic field is 0.0315 \(\mathrm{T}\) . The ions next enter a uniform magnetic field of magnitude 0.0175 \(\mathrm{T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{cm},\) what is their mass?

. You have 25 \(\mathrm{m}\) of wire, which you want to use to construct a 44 \(\mathrm{cm}\) diameter coil whose magnetic field at its center will exactly cancel the earth's field of 0.55 gauss. What current will your coil require?

A long, straight telephone cable contains six wires, each carrying a current of 0.300 A. The distances between wires can be neglected. (a) If the currents in all six wires are in the same direction, what is the magnitude of the magnetic field 2.50 \(\mathrm{m}\) from the cable? (b) If four wires carry currents in one direction and the other two carry currents in the opposite direction, what is the magnitude of the field 2.50 m from the cable?

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