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Chapter 19: Problem 58

A beam of protons is accelerated easterly from rest through a potential difference of \(3.0 \mathrm{kV}\). It enters a region where there exists an upward pointing uniform electric field. This field is created by two parallel plates separated by \(10 \mathrm{~cm}\) with a potential difference of \(250 \mathrm{~V}\) across them. (a) What is the speed of the protons as they enter the electric field? (b) Find the magnitude and direction relative to the velocity of the magnetic field (perpendicular to \(\overrightarrow{\mathrm{E}})\) needed so the beam passes undeflected through the plates. (c) What happens to the protons if the magnetic field is greater than the value found in part (b)?

Short Answer

Expert verified
(a) Speed is \(2.39 \times 10^5\) m/s. (b) \(B = 0.0105 \text{ T}\). (c) Protons deflect downward if \(B\) is greater.

Step by step solution

01

Identify Energy Conversion

The protons start from rest, so their initial kinetic energy is zero. The energy gained by the protons when they are accelerated through a potential difference is given by the work-energy principle where the gain in kinetic energy equals the loss in potential energy.
02

Calculate Speed of Protons

The kinetic energy acquired by the protons can be expressed as \( KE = \frac{1}{2}mv^2 = qV \), where \( q \) is the charge of a proton, \( m \) is its mass, \( V \) is the accelerating potential difference, and \( v \) is the speed of the protons. Rearranging for \( v \), we have: \[ v = \sqrt{\frac{2qV}{m}} \]. Using the values \( q = 1.6 \times 10^{-19} \text{ C}, \ m = 1.67 \times 10^{-27} \text{ kg}, \) and \( V = 3000 \text{ V} \), calculate \( v \) as follows: \[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 3000}{1.67 \times 10^{-27}}} = 2.39 \times 10^5 \text{ m/s} \].
03

Calculate the Electric Field Between Plates

The electric field \( E \) between the plates is \( E = \frac{V}{d} \), where \( V = 250 \text{ volts} \) and \( d = 0.1 \text{ m} \). Calculate \( E \) as:\[ E = \frac{250}{0.1} = 2500 \text{ N/C} \].
04

Identify Magnetic Field to Prevent Deflection

To ensure the protons pass through the field undeflected, the magnetic force \( qvB \) must equal the electric force \( qE \). Therefore, \( qvB = qE \), implying \( vB = E \). Rearranging for \( B \), the magnetic field is \( B = \frac{E}{v} \). Calculate \( B \) using \( E = 2500 \text{ N/C} \) and \( v = 2.39 \times 10^5 \text{ m/s} \). Thus, \[ B = \frac{2500}{2.39 \times 10^5} \approx 0.0105 \text{ T} \].
05

Analyze the Effect of Greater Magnetic Field

If the magnetic field is greater than 0.0105 T, the magnetic force \( qvB \) will exceed the electric force \( qE \), resulting in the protons being deflected downward, assuming the correct direction perpendicular to velocity.

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Key Concepts

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

Proton Acceleration
Proton acceleration involves increasing the speed of protons from rest using an electric potential difference. In our context, protons are particles with a positive charge that, when exposed to an electric field, experience a force that accelerates them.

The potential difference of 3.0 kV (or 3000 volts) applied to the protons serves as the energy source for their acceleration. This potential difference accelerates the protons in the electric field, converting potential energy to kinetic energy. We use the formula for kinetic energy resulting from an electric potential, given by:
  • The work-energy principle: \[ KE = rac{1}{2}mv^2 = qV \]
Here, \( KE \) represents the kinetic energy, \( m \) is the mass of the proton, \( v \) is its velocity, \( q \) is the charge of the proton, and \( V \) is the potential difference.

As the protons start from rest, their initial kinetic energy is zero. Therefore, all energy from the potential difference is transformed into kinetic energy. Upon calculation with given values, protons achieve a speed of approximately \( 2.39 \times 10^5 \) m/s when they enter the electric field. Understanding this conversion helps clarify the acceleration mechanism faced by charged particles like protons.
Potential Difference
The potential difference, often called voltage, is a key element in the movement of charged particles such as protons in an electric field. It indicates the difference in electric potential energy per unit charge between two points in an electric field. This potential difference is responsible for exerting a force on charges, giving them kinetic energy, as observed in protons here.

When protons move through a potential difference of 3.0 kV, they gain energy which is converted into their kinetic form, allowing them to accelerate. The potential difference impacts the magnitude of the acceleration; higher potential differences result in faster speeds.

Similarly, within the region containing two parallel plates, a smaller potential difference of 250 V creates an electric field between them. This field exerts forces on the protons, affecting their trajectory unless other forces (such as magnetic forces) balance them. This concept illustrates how voltage not only governs speed but also the direction of charged particles.
Undeflected Motion
Undeflected motion refers to a situation where a particle moves through a field without deviation from its original path. In our exercise, we want the protons to pass straight through the electric field between the plates without bending.

To achieve this, we exploit the creation of a magnetic field perpendicular to the electric field. The electric field applies a force upward on the protons, while a suitable magnetic field downwards can counterbalances this force.

This is accomplished by setting the magnetic force equal to the electric force, utilizing the equation:
  • \[ qvB = qE \] leading to \( vB = E \)
Ultimately, the magnetic field value required is calculated using the velocity of the protons and the electric field strength, resulting in approximately 0.0105 T to ensure no deflection. If the magnetic field exceeds this value, it would overpower the electric force, leading to a deflected trajectory for the protons. Understanding this balance between electric and magnetic forces is crucial to controlling particle paths in a variety of scientific and practical applications.

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Most popular questions from this chapter

A charge of \(0.250 \mathrm{C}\) moves vertically in a field of \(0.500 \mathrm{~T}\) that is oriented some angle from the vertical. If the charge's speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s}\), what field angle(s) will ensure that the force acting on the charge is \(5.0 \mathrm{~N}\) ?

Exiting a linear accelerator, a narrow horizontal beam of protons travels due north. If \(1.75 \times 10^{13}\) protons pass a given point per second, (a) determine the magnetic field direction and strength at a location of \(2.40 \mathrm{~m}\) east of the beam. (b) Does it seem likely this would demagnetize the encoded magnetic strip on, for example, an ATM card? [Hint: The ATM card "lives" safely in the Earth's magnetic environment.]

(a) What angle(s) does a particle's velocity have to make with the magnetic field direction for the particle to be subjected to half the maximum possible magnetic force, \(F_{\max } ?\) (b) Express the magnetic force on a charged particle in terms of \(F_{\max }\) if the angle between its velocity and the magnetic field direction is (i) \(10^{\circ},\) (ii) \(80^{\circ},\) and (iii) \(100^{\circ} .\) (c) If the particle's velocity makes an angle of \(50^{\circ}\) with respect to the magnetic field direction, at what other angle(s) would the magnetic force on it be the same? Would the direction be the same? Explain.

A wire carries a current of \(10 \mathrm{~A}\) in the \(+x\) -direction. (a) Find the force per unit length on the wire if it is in a magnetic field that has components of \(B_{x}=0.020 \mathrm{~T}\) \(B_{y}=0.040 \mathrm{~T},\) and \(B_{z}=0 \mathrm{~T} .\) (b) Find the force per unit length on the wire if only the field's \(x\) -component is changed to \(B_{x}=0.050 \mathrm{~T}\). (c) Find the force per unit length on the wire if only the field's \(y\) -component is changed to \(B_{y}=-0.050 \mathrm{~T}\)

A cylindrical solenoid \(10 \mathrm{~cm}\) long has 3000 turns of wire and carries a current of 5.0 A. A second solenoid, consisting of 2000 turns of wire and the same length as the first solenoid, surrounds it and is concentric (shares a common central axis) with it. The outer coil carries a current of 10 A in the same direction as the current in inner one. (a) Find the magnetic field near their common center. (b) What current in the second solenoid (magnitude and relative direction) would make the net field strength at the center twice that of the first solenoid alone? (c) What current in the second solenoid (magnitude and relative direction) would result in zero net magnetic field near their common center?
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