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An electron passes through a magnetic field without being deflected. What do you conclude about the magnetic field?

Short Answer

Expert verified
If an electron passes through a magnetic field without being deflected, we can conclude that either the magnetic field is zero or that the magnetic field is parallel or antiparallel to the velocity of the electron. In both scenarios, the magnetic force on the electron is zero, leading to no deflection.

Step by step solution

01

Recall the formula for magnetic force

The magnetic force experienced by a charged particle (like an electron) is given by the Lorentz force formula: \[F = q(\textbf{v} \times \textbf{B})\] where \(F\) is the magnetic force, \(q\) is the charge of the particle, \(\textbf{v}\) is the velocity of the particle, and \(\textbf{B}\) is the magnetic field vector. The cross product (\(\times\)) indicates that the force is perpendicular to both the velocity and magnetic field vectors.
02

Identify the condition for zero magnetic force

For the magnetic force to be zero, it means that the cross product between \(\textbf{v}\) and \(\textbf{B}\) should be zero. This can occur in two scenarios: 1. When the magnetic field \(\textbf{B}\) is zero. 2. When the magnetic field \(\textbf{B}\) is parallel or antiparallel to the velocity vector \(\textbf{v}\) (meaning that they are in the same or opposite directions). In this case, their cross product will be zero as the sine of the angle between them will be zero.
03

Conclusion

If an electron passes through a magnetic field without being deflected, we can conclude that either the magnetic field is zero or that the magnetic field is parallel or antiparallel to the velocity of the electron. In both scenarios, the magnetic force on the electron is zero, leading to no deflection.

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Key Concepts

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

Lorentz Force
The Lorentz force is a fundamental concept in physics, describing the force exerted on a charged particle moving in a magnetic field. It is given by the equation: \[F = q(\textbf{v} \times \textbf{B})\]where \(F\) represents the force on the particle, \(q\) is the charge of the particle, \(\textbf{v}\) is the velocity vector, and \(\textbf{B}\) is the magnetic field vector. The direction of the force is determined by the right-hand rule, which finds a vector perpendicular to the plane formed by the velocity and the magnetic field vectors. For students to really grasp this concept, it's helpful to think of \(\textbf{v}\) and \(\textbf{B}\) as arrows. If the arrows are aligned or exactly opposite, they can't 'push' on each other, leading to no perpendicular 'push' or force. Thus, the electron path remains undeflected if the magnetic field is zero or parallel to its velocity. This principle underlies devices from electric motors to particle accelerators.
Magnetic Field Vector
The magnetic field vector, denoted by \(\textbf{B}\), represents the strength and direction of a magnetic field in a given point in space. It is a vector quantity, which means it has both magnitude (how strong the field is) and direction (where the field points). In the context of our exercise, if an electron moves through a magnetic field without being deflected, this tells us that the direction of the \(\textbf{B}\) vector can be deduced - it must be aligned with the velocity \(\textbf{v}\) of the electron. Understanding magnetic fields is crucial in contexts ranging from designing compasses to understanding how galaxies form.
Cross Product in Physics
The cross product is an operation that takes two vectors and returns a third vector that is perpendicular to the plane containing the original vectors. It's represented mathematically as \(\textbf{A} \times \textbf{B}\). The magnitude of the resultant vector is given by \(|\textbf{A}| |\textbf{B}| \sin(\theta)\), where \(|\textbf{A}|\) and \(|\textbf{B}|\) are the magnitudes of the original vectors, and \(\theta\) is the smallest angle between them. When learning about the cross product, visualize your right hand with your index finger pointing in the direction of the first vector (\(\textbf{A}\)) and your middle finger in the direction of the second vector (\(\textbf{B}\)); your thumb will then point in the direction of the cross product result. In our exercise scenario, since the electron is not deflected, it implies that the cross product of its velocity and the magnetic field must be zero, showing either parallelism or a zero magnetic field.
Charge Particle Behavior in Magnetic Field
Charged particles, such as electrons, experience forces when they move through magnetic fields, leading to changes in their trajectories. This behavior is governed by the Lorentz force and if the particle's velocity is perpendicular to the magnetic field, it will follow a circular or helical path. However, if the velocity is parallel or antiparallel to the magnetic field, as the exercise suggests, there is no force and thus no deflection. This behavior allows us to manipulate charged particles for use in applications like cathode ray tubes or magnetic resonance imaging (MRI) machines. When visualizing this concept, imagine the path of the particle bending only when the magnetic 'wind' can 'push' on it from the side, not when it moves with or against the flow.

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Most popular questions from this chapter

(a) An oxygen-16 ion with a mass of $$2.66 \times 10^{-26} \mathrm{kg} \text { travels at } 5.0 \times 10^{6} \mathrm{m} / \mathrm{s}$$ perpendicular to a 1.20 -T magnetic field, which makes it move in a circular arc with a 0.231 -m radius. What positive charge is on the ion? (b) What is the ratio of this charge to the charge of an electron? (c) Discuss why the ratio found in (b) should be an integer.

At a given instant, an electron and a proton are moving with the same velocity in a constant magnetic field. Compare the magnetic forces on these particles. Compare their accelerations.

A circular coil with 200 turns has a radius of 2.0 cm. (a) What current through the coil results in a magnetic dipole moment of \(3.0 \mathrm{Am}^{2}\) ? (b) What is the maximum torque that the coil will experience in a uniform field of strength \(5.0 \times 10^{-2} \mathrm{T} ?\) (c) If the angle between \(\mu\) and \(B\) is \(45^{\circ},\) what is the magnitude of the torque on the coil? (d) What is the magnetic potential energy of coil for this orientation?

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a \(0.500-\mu C\) charge and flies due west at a speed of \(660 .\) m/s over Earth's south magnetic pole, where the \(8.00 \times 10^{-5}-\mathrm{T}\) magnetic field points straight down into the ground. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.

Calculate the magnetic field strength needed on a 200-turn square loop 20.0 cm on a side to create a maximum torque of \(300 \mathrm{N} \cdot \mathrm{m}\) if the loop is carrying 25.0 A.

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