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An electron travels in the \(+x\) -direction in a magnetic field and is acted on by a magnetic force in the \(-y\) -direction. (a) In which of the following directions could the magnetic field be oriented: \((1)-x,(2)+y\) (3) \(+z,\) or \((4)-z ?\) Explain. \((b)\) If the electron speed is \(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) and the magnitude of the force is \(5.0 \times 10^{-19} \mathrm{~N},\) what is the magnetic field strength?

Short Answer

Expert verified
(a) The magnetic field direction is (+z). (b) The magnetic field strength is about 1.04 x 10^-4 T.

Step by step solution

01

Understand the Direction of Magnetic Force

The electron is moving in the positive x-direction and the magnetic force is acting in the negative y-direction. According to the right-hand rule and given that the charge of an electron is negative, the magnetic field direction should be such that the force points in the -y direction as per the left-hand rule (or the reverse of the right-hand rule for positive particles).
02

Determine the Possible Magnetic Field Direction

The possible direction of the magnetic field can be determined using the vector cross-product \[\vec{F} = q(\vec{v} \times \vec{B})\] which indicates that the force \(\vec{F}\) is perpendicular to both the velocity \(\vec{v}\) and the magnetic field \(\vec{B}\). If velocity \(\vec{v}\) is along +x and force \(\vec{F}\) is along -y, then \(\vec{B}\) should be along +z to have the magnetic force act in the -y direction with respect to an electron's motion.
03

Apply the Formula to Calculate Magnetic Field Strength

We use the formula for the magnetic force \[F = qvB\sin\theta\]. Given \(F = 5.0 \times 10^{-19} \text{ N}\), \(v = 3.0 \times 10^6 \text{ m/s}\), \(q = 1.6 \times 10^{-19} \text{ C}\) and \(\theta = 90^\circ\) (since the force and velocity are perpendicular), the magnetic field strength \(B\) is calculated as: \[B = \frac{F}{qv} = \frac{5.0 \times 10^{-19}}{1.6 \times 10^{-19} \times 3.0 \times 10^6} \approx 1.04 \times 10^{-4} \text{ T}\].

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Key Concepts

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

Electromagnetic Theory
Electromagnetic theory studies how electric and magnetic fields interact and affect objects. This interaction is crucial in understanding forces experienced by charged particles, like electrons, moving through magnetic fields. An electron moving through a magnetic field is influenced by the magnetic forces acting on it. These forces do not work in isolation but depend on the field’s strength and direction, as well as the speed and direction of the electron itself. The magnetic force on a charged particle is described by several factors:
  • The charge of the particle: Electrons are negatively charged.
  • The velocity of the particle: Determines the magnitude of the force experienced.
  • The strength and direction of the magnetic field: Dictates the force vector.
When electrons travel through a uniform magnetic field, they experience a perpendicular force that causes them to curve. This occurrence reflects the core importance of electromagnetic theories in predicting particle trajectories.
Vector Cross-Product
The vector cross-product is a mathematical operation crucial in physics, particularly in analyzing forces like the magnetic force on a charged particle. The formula \[ \vec{F} = q(\vec{v} \times \vec{B}) \] explains how this works. Here, \( \vec{F} \) is the magnetic force, \( q \) is the charge of the particle, \( \vec{v} \) is its velocity, and \( \vec{B} \) is the magnetic field.This operation results in a vector (\( \vec{F} \)) that is perpendicular to both the velocity (\( \vec{v} \)) and the magnetic field (\( \vec{B} \)). For example, if an electron is traveling in the \(+x\) direction and the force is observed in the \(-y\) direction, the magnetic field should be oriented in the \(+z\) direction to satisfy these conditions.Key characteristics of the vector cross-product include:
  • It results in a vector perpendicular to the original vectors.
  • Its magnitude depends on the sine of the angle between the vectors \(\theta\), reaching its maximum when the vectors are perpendicular (\(\theta = 90^\circ\)).
  • It plays a crucial role in determining the three-dimensional orientation of forces in fields such as physics and engineering.
Right-Hand Rule
The right-hand rule is a simple yet powerful tool to determine the direction of vectors resulting from a cross-product. For positive charges, align your right hand in the direction of the first vector (velocity), curl your fingers towards the second vector (magnetic field), and your thumb will point in the direction of the resulting vector (force).However, since electrons are negatively charged, the rule's application reverses. This is equivalent to using the left-hand rule for electrons. In our specific exercise, as the electron travels in the \(+x\) direction:
  • Velocity \( \vec{v} \) is in the \(+x\) direction.
  • Force \( \vec{F} \) is in the \(-y\) direction.
  • Thus, the magnetic field \( \vec{B} \) must be in the \(+z\) direction to satisfy these conditions.
Using either the right-hand or left-hand rule, depending on the charge, helps predict the orientation of vectors easily during calculations. This practical approach is extensively used in electromagnetic applications to gain accurate insights into vector dynamics and orientations.

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Most popular questions from this chapter

A circular coil of current-carrying wire has the normal to its area pointing upward. A second smaller concentric circular coil carries a current in the opposite direction. (a) Where, in the plane of these coils, could the magnetic field be zero: (1) only inside the smaller one, (2) only between the inner and outer one, (3) only outside the larger one, or (4) inside the smaller one and outside the larger one? (b) The larger one is a 200-turn coil of wire with a radius of \(9.50 \mathrm{~cm}\) and carries a current of \(11.5 \mathrm{~A}\). The second one is a 100 -turn coil with a radius of \(2.50 \mathrm{~cm}\). Determine the current in the inner coil so the magnetic field at their common center is zero. Neglect the Earth's field.

A charged particle travels undeflected through perpendicular electric and magnetic fields whose magnitudes are \(3000 \mathrm{~N} / \mathrm{C}\) and \(30 \mathrm{mT}\), respectively. Find the speed of the particle if it is (a) a proton and (b) an alpha particle. (An alpha particle is a helium nucleus-a positive ion with a double positive charge of \(+2 e .\) )

A particle with charge \(q\) and mass \(m\) moves in a horizontal plane at right angles to a uniform vertical magnetic field \(B\). (a) What are the period \(T\) and frequency \(f\) of the particle's circular motion in terms of \(q, B\), and \(m\) ? (This frequency is called the cyclotron frequency.) You result should verify that the time for one orbit for any charged particle in a uniform magnetic field is independent of its speed and radius. (b) Compute the path radius and the cyclotron frequency if the particle is an electron with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\) traveling in a region where the field strength is \(0.10 \mathrm{mT}\)

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 long, straight wire carries a current of \(2.5 \mathrm{~A}\). Find the magnitude of the magnetic field \(25 \mathrm{~cm}\) from the wire.

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