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An electron is moving along positive \(x\) -axis. To get it moving on an anticlockwise circular path in \(x-y\) plane, a magnetic field is applied a along positive \(y\) -axis b. along positive \(z\) -axis c. along negative \(y\) -axis d. along negative \(z\) -axis

Short Answer

Expert verified
b. Along positive z-axis.

Step by step solution

01

Identify the Force Direction with Right-Hand Rule

To determine the path of the electron, we use the right-hand rule for magnetic forces. When a positive charge moves through a magnetic field, the force is perpendicular to both the velocity and the magnetic field. However, for an electron, the force direction is opposite, because it has a negative charge.
02

Analyze the Situation for Each Magnetic Field Direction

Since the electron moves along the positive x-axis, consider each option: - a. Magnetic field is along positive y-axis: For a positive charge, force would be along positive z; for an electron, it is along negative z. - b. Magnetic field is along positive z-axis: For a positive charge, force would be along negative y; for an electron, it is along positive y. - c. Magnetic field is along negative y-axis: For a positive charge, force would be along negative z; for an electron, it is along positive z. - d. Magnetic field is along negative z-axis: For a positive charge, force would be along positive y; for an electron, it is along negative y.
03

Determine the Anticlockwise Circular Path

The electron must experience a centripetal force directed towards the center of the circular path. When the magnetic field is along the positive z-axis, the force on the electron is along positive y-axis, providing the right conditions for movement in an anticlockwise direction in the x-y plane.

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Key Concepts

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

Right-Hand Rule
The right-hand rule is a helpful tool in understanding directions in electromagnetism, especially when it comes to magnetic forces and fields. For situations involving charged particles in a magnetic field, the right-hand rule allows you to determine the direction of the force acting upon the charge.

Here's how it works: to determine the force direction on a positive charge, align your right-hand fingers in the direction of its velocity. Then, rotate your wrist until the fingers point in the direction of the magnetic field. Your thumb will point in the direction of the force. However, since an electron is a negative charge, the force direction is opposite to what's indicated by your thumb.

It's crucial to understand this inversion for electrons.
  • When the force appears to be upwards for a positive charge, it will be downwards for an electron.
  • If the force points north for a positive charge, it will point south for an electron.
This directional understanding is foundational when working through problems involving electron motion in a magnetic field.
Centripetal Force
Centripetal force is at the heart of understanding why an electron can move in a circular path. Essentially, this force acts towards the center of a circular path and is responsible for maintaining a particle's circular motion. In the case of an electron traveling through a magnetic field, the magnetic force itself acts as this centripetal force.

For a charge moving at a velocity \( v \) in a magnetic field, the magnetic force \( F_B \) is given by:
\[ F_B = qvB \sin \theta \]
where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.

For circular motion, \( \theta \) is typically \( 90^\circ \), simplifying \( \sin \theta = 1 \). Thus, the force maintains a constant perpendicular direction to the velocity, pulling the particle into a curved trajectory.

This centripetal force must be matched perfectly to enable the electron to move in a stable circular motion without changing its speed.
Electron Motion
When discussing electron motion in a magnetic field, it's vital to revisit how the forces interact to produce the observed paths. As detailed through the use of the right-hand rule, knowing the direction of force is essential. For electrons, which are negatively charged, the force they experience in a magnetic field results in a path opposite that of a positive charge under similar conditions.

The magnetic force acting on an electron traveling in the \( x-y \) plane can induce circular motion. This circular path happens because the magnetic force is always perpendicular to the velocity of the electron. Hence, instead of accelerating or decelerating the electron, it alters the direction of its motion, resulting in a circular trajectory.

Understanding this principle of electron motion is crucial for predicting how they behave in magnetic fields. Various applications harness this behavior, such as in cyclotrons and other particle accelerators, where electrons move in controlled circular paths for research and practical applications. Analyzing these motions not only explains theoretical principles but also underpins many practical technologies in physics and engineering.

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Most popular questions from this chapter

Two parallel wires carrying equal currents in opposite directions are placed at \(x=\pm a\) parallel to \(y\) -axis with \(z=0\). Magnetic field at origin \(O\) is \(B_{1}\) and at \(P(2 a, 0,0)\) is \(B_{2}\). Then, the ratio \(B_{1} / B_{2}\) is a. \(-3\) b. \(-\frac{1}{2}\) c. \(-\frac{1}{3}\) d. 2

A loop of flexible conducting wire of length \(\ell\) lies in magnetic field \(B\) which is normal to the plane of loop. A current \(I\) is passed through the loop. The tension developed in the wire to open up is a. \(\frac{\pi}{2} B I \ell\) b. \(\frac{B I \ell}{2}\) c. \(\frac{B l \ell}{2 \pi}\) d. BI\ell

Two parallel wires carry currents of 20 and \(40 \mathrm{~A}\) in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be a toward \(20 \mathrm{~A} \quad \mathrm{~b}\) toward \(40 \mathrm{~A}\) c. zero d perpendicular to the plane of the currents

Two identical wires \(A\) and \(B\) have the same length \(\ell\) and carry the same current \(I .\) Wire \(A\) is bent into a circle of radius \(R\) and wire \(B\) is bent to form a square of side \(a\). If \(B_{1}\) and \(B_{2}\) are the values of magnetic induction at the center of the circle and the centre of the square, respectively, then the ratio \(B_{1} / B_{2}\) is a. \(\left(n^{2} / 8\right)\) h \(\left(\pi^{2} / 8 \sqrt{2}\right)\) c. \(\left(n^{2} / 16\right)\) d. \(\left(\pi^{2} / 16 \sqrt{2}\right)\)

A particle of charge per unit mass \(\alpha\) is released from origin with velocity \(\vec{v}=v_{0} \hat{i}\) in a magnetic field $$ \bar{B}=-B_{0} \hat{k} \quad \text { for } x \leq \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha} $$ and \(\bar{B}=0\) $$ \text { for } x>\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha} $$ The \(x\) -coordinate of the particle at time \(t\left(>\frac{\pi}{3 B_{0} \alpha}\right)\) would be a. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{\sqrt{3}}{2} v_{0}\left(t-\frac{\pi}{B_{0} \alpha}\right)\) b \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+v_{0}\left(t-\frac{\pi}{3 B_{0} \alpha}\right)\) c. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{v_{0}}{2}\left(t-\frac{\pi}{3 B_{0} \alpha}\right)\) d. \(\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} \alpha}+\frac{v_{0} t}{2}\)

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