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Chapter 27: Problem 13

Determine the magnitude and direction of the force on an electron traveling \(8.75 \times 10^{5} \mathrm{~m} / \mathrm{s}\) horizontally to the east in a vertically upward magnetic field of strength \(0.45 \mathrm{~T}\).

Short Answer

Expert verified
The magnitude of the force is \( 6.3 \times 10^{-14} \text{ N} \) directed south.

Step by step solution

01

Understand the Problem

This problem involves an electron moving in a magnetic field. We need to find the magnitude and direction of the force acting on it. The electron moves horizontally to the east with a velocity of \( 8.75 \times 10^{5} \text{ m/s} \) in a magnetic field oriented vertically upwards with a strength of \( 0.45 \text{ T} \).
02

Identify the Formula

The force on a charged particle moving in a magnetic field is given by the Lorentz force formula: \[ F = qvB \sin\theta \] where \( F \) is the force, \( q \) is the charge of the particle, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. Here, \( \theta = 90^\circ \) since the velocity is perpendicular to the magnetic field.
03

Calculate the Charge of the Electron

An electron has a charge of \( -1.6 \times 10^{-19} \text{ C} \).
04

Apply the Lorentz Force Formula

Since the velocity and the magnetic field are perpendicular, \( \sin 90^\circ = 1 \). Therefore, the force is: \[ F = qvB = (-1.6 \times 10^{-19} \text{ C})(8.75 \times 10^{5} \text{ m/s})(0.45 \text{ T}) \] \[ F = -6.3 \times 10^{-14} \text{ N} \] The negative sign indicates the direction of the force.
05

Determine the Direction of the Force

The direction of the force can be determined using the right-hand rule. For an electron (negative charge) moving to the east in an upward magnetic field, the force is directed to the south. The negative sign indicates that the force direction is opposite to what the right-hand rule gives for a positive charge.

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible field that exerts force on moving charged particles, such as electrons. It can be visualized as a series of lines, with direction pointing from the north pole to the south pole, indicating the path of the force. Magnetic fields are characterized by their strength and direction. In this exercise, the magnetic field is stated to have a strength of \(0.45 \text{ T}\) and points vertically upward. This information is critical, as the interaction between the electron's motion and this field determines the force exerted on the electron.

  • Units: The unit of magnetic field strength is the Tesla (\( \text{T} \)).
  • Direction: Upward, which affects how forces act on charged particles.
Electron Charge
The electron is a subatomic particle carrying a negative charge, denoted by \(-1.6 \times 10^{-19} \text{ C}\). This charge is fundamental to calculating the force exerted by a magnetic field. The negative sign plays a crucial role when applying the Lorentz force equation, which calculates the force experienced by a charged particle moving through a magnetic field. The charge magnitude does not change, but its negative sign indicates that the force direction will differ from that of a positively charged particle. This becomes particularly relevant when using rules like the right-hand rule to determine force direction.

  • Charge value: \(-1.6 \times 10^{-19} \text{ C}\).
  • Impact of charge: Negative charge affects direction calculations using the right-hand rule.
Right-Hand Rule
The right-hand rule is a mnemonic used to determine the direction of the force on a charged particle moving in a magnetic field. For a positively charged particle, point the thumb of your right hand in the direction of velocity, your fingers in the direction of the magnetic field, and the force direction will come out of your palm. However, because electrons have a negative charge, the direction given by the right-hand rule must be reversed. In this exercise, since the electron moves east and the magnetic field goes upward, applying the right-hand rule yields a southward force direction for a positive charge. For the electron's negative charge, the actual force direction is opposite, to the north.

  • Step-by-step: Thumb = velocity (east), Fingers = magnetic field (up).
  • For negatives: Flip the direction given by the palm for an electron's force.
Force Direction
The direction of the force experienced by a charged particle moving in a magnetic field is determined by both the right-hand rule and the sign of the charge. For this exercise, the electron's negative charge results in a force direction that is the reverse of the direction given by the right-hand rule for positive charges. The calculation revealed a southward force for a positive charge, but because electrons are negatively charged, the final force direction is towards the north. Understanding this reversal is key in correctly interpreting the problem's outcome.

  • Result for positive: South (initial right-hand rule direction).
  • Actual for electron: North (due to negative charge reversal).

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Most popular questions from this chapter

The force on a wire is a maximum of \(7.50 \times 10-2 \mathrm{~N}\) when placed betw een the pole faces of a magnet. The current flows horizontally to the right and the magnetic field is vertical. The wire is observed to "jump" toward the observer when the current is turned on. (a) What type of magnetic pole is the top pole face? (b) If the pole faces have a diameter of \(10.0 \mathrm{~cm},\) estimate the current in the wire if the field is \(0.220 \mathrm{~T}\) (c) If the wire is tipped so that it makes an angle of \(10.0^{\circ}\) with the horizontal, what force will it now feel?

A Hall probe used to measure magnetic field strengths consists of a rectangular slab of material (free-electron density \(n\) ) with width \(d\) and thickness \(t,\) carrying a current \(I\) along its length \(\ell\). The slab is immersed in a magnetic field of magnitude \(B\) oriented perpendicular to its rectangular face (of area \(\ell d\) ), so that a Hall emf \(\mathscr{E}_{\mathrm{H}}\) is produced across its width \(d\). The probe's magnetic sensitivity, defined as \(K_{\mathrm{H}}=\mathscr{E}_{\mathrm{H}} / I B,\) indicates the magnitude of the Hall emf achieved for a given applied magnetic field and current. A slab with a large \(K_{\mathrm{H}}\) is a good candidate for use as a Hall probe. (a) Show that \(K_{\mathrm{H}}=1\) /ent. Thus, a good Hall probe has small values for both \(n\) and \(t .(b)\) As possible candidates for the material used in a Hall probe, consider (i) a typical metal \(\left(n \approx 1 \times 10^{29} / \mathrm{m}^{3}\right)\) and (ii) a (doped) semiconductor \(\left(n \approx 3 \times 10^{22} / \mathrm{m}^{3}\right)\). Given that a semiconductor slab can be manufactured with a thickness of \(0.15 \mathrm{~mm}\), how thin (nm) should a metal slab be to yield a \(K_{H}\) value equal to that of the semiconductor slab? Compare this metal slab thickness with the 0.3 -nm size of a typical metal atom. (c) For the typical semiconductor slab described in part \((b),\) what is the expected value for \(\mathscr{E}_{\mathrm{H}}\) when \(I=100 \mathrm{~mA}\) and \(B=0.1 \mathrm{~T} ?\)

(II) A doubly charged helium atom whose mass is \(6.6 \times 10^{-27} \mathrm{kg}\) is accelerated by a voltage of 2700 \(\mathrm{V}\) . (a) What will be its radius of curvature if it moves in a plane perpendicular to a uniform \(0.340-\mathrm{T}\) field? (b) What is its period of revolution?

In a probe that uses the Hall effect to measure magnetic fields, a 12.0-A current passes through a 1.50 -cm-wide 1.30-mm-thick strip of sodium metal. If the Hall emf is \(1.86 \mu \mathrm{V},\) what is the magnitude of the magnetic field (take it perpendicular to the flat face of the strip)? Assume one free electron per atom of \(\mathrm{Na}\), and take its specific gravity to be 0.971 .

A proton follows a spiral path through a gas in a magnetic field of \(0.018 \mathrm{~T}\), perpendicular to the plane of the spiral, as shown in Fig. \(27-54 .\) In two successive loops, at points \(P\) and \(\mathrm{Q},\) the radii are \(10.0 \mathrm{~mm}\) and \(8.5 \mathrm{~mm},\) respectively. Calculate the change in the kinetic energy of the proton as it travels from \(P\) to \(Q\)
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