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An electron has velocity \(\vec{v}=(32 \hat{i}+40 \hat{j}) \mathrm{km} / \mathrm{s}\) as it enters a uniform magnetic field \(\vec{B}=60 \hat{\mathrm{i}} \mu \mathrm{T}\). What are (a) the radius of the helical path taken by the electron and (b) the pitch of that path? (c) To an observer looking into the magnetic field region from the entrance point of the electron, does the electron spiral clockwise or counterclockwise as it moves?

Short Answer

Expert verified
(a) The radius is approximately 0.038 m, (b) the pitch is approximately 1.91 μm, (c) the electron spirals counterclockwise.

Step by step solution

01

Determine the Perpendicular Velocity Component

The velocity component perpendicular to the magnetic field is solely in the \(\hat{j}\) direction, as \(\vec{B}=60 \hat{\mathrm{i}}\). Thus, \(v_\perp = 40 \, \text{km/s} = 40\times10^3 \, \text{m/s}\).
02

Calculate the Radius of the Helical Path

The radius \( r \) of an electron in a magnetic field can be calculated using the formula \( r = \frac{m v_\perp}{q B} \), where \( m = 9.11 \times 10^{-31} \, \text{kg}\), \( q = 1.6 \times 10^{-19} \, \text{C}\), and \( B = 60 \times 10^{-6} \, \text{T}\). Substituting the values, we find \( r = \frac{9.11 \times 10^{-31} \times 40 \times 10^3}{1.6 \times 10^{-19} \times 60 \times 10^{-6}} \approx 3.80 \times 10^{-2} \, \text{m}\).
03

Determine the Parallel Velocity Component

Since the component of the velocity parallel to the magnetic field doesn't change, \(v_{||} = 32 \, \text{km/s} = 32 \times 10^3 \, \text{m/s}\).
04

Calculate the Pitch of the Helical Path

The pitch \( p \) of the helical path is given by \( p = v_{||} \cdot T \), where \( T \) is the period of rotation given by \( T = \frac{2\pi m}{qB} \). Thus, \( T = \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 60 \times 10^{-6}} \approx 5.97 \times 10^{-11} \, \text{s}\), and \( p = 32 \times 10^3 \times 5.97 \times 10^{-11} \approx 1.91 \times 10^{-6} \, \text{m}\).
05

Determine Spiral Direction

Using the right-hand rule, point your thumb in the direction of \(\vec{B}\) and your fingers in the direction of \(\vec{v}_\perp\). Your fingers will curl in the counterclockwise direction, indicating the electron spirals counterclockwise as seen from the entrance point.

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Key Concepts

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

Helical Path
When an electron enters a magnetic field at an angle, it doesn't just move in a straight line or simple circle. Instead, it follows a helical, or spiral-like, path. This happens because part of the electron's velocity is directed along the magnetic field lines, while the rest is perpendicular. The component of velocity parallel to the magnetic field lets the electron keep moving forward. Meanwhile, the perpendicular component causes it to circle around the magnetic lines.
  • Parallel motion keeps the electron moving along the field.
  • Perpendicular motion creates circular orbits around the field lines.
Together, these motions create a helical or spiral path. By calculating both velocity components, we can find the radius of this helical path. This helps describe how tightly the electron spirals as it moves through the field.
Electron Motion
Electrons have a negative charge, meaning that they're inherently responsive to magnetic fields. When an electron with velocity \((32 \hat{i}+40 \hat{j}) \, \text{km/s}\) enters a magnetic field \((60 \hat{i} \, \mu \text{T})\), we can break down its motion into parts.
  • The perpendicular velocity component, calculated as \(40 \, \text{km/s} = 40 \times 10^3 \, \text{m/s}\), governs the circular motion.
  • The parallel velocity component, \(32 \, \text{km/s} = 32 \times 10^3 \, \text{m/s}\), ensures linear forward motion.
The magnetic field doesn't change the speed of the electron but redirects it into a spiral. By analyzing these motions, the helix's radius and pitch (or spiral step distance) can be determined.>The radius is determined by the mass, charge, and perpendicular velocity of the electron as well as the strength of the magnetic field.>The pitch is derived from the period of rotation and the parallel velocity. Understanding these aspects shows how the electron's path morphs into a helix.
Right-Hand Rule
Understanding the direction of electron motion in a magnetic field involves a handy tool known as the right-hand rule. Although the right-hand rule is typically applied to positively charged particles, for electrons, which are negatively charged, the rule is applied inversely.To use the right-hand rule:
  • Point your thumb in the direction of the magnetic field \(\vec{B}\).
  • Let your fingers follow the direction of the electron's perpendicular velocity \(\vec{v}_\perp\).
  • Your palm's direction then indicates the electron's force, guiding the spiral motion.
In this specific problem, pointing your right thumb along \(\hat{i}\) and curling your fingers along \(\hat{j}\), determines that the spiral, as viewed from the entrance, moves counterclockwise. This intuitive method simplifies visualizing particle movements in a magnetic field, reinforcing the concept that the magnetic force acts perpendicular to both the velocity and field.

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Most popular questions from this chapter

A wire \(1.80 \mathrm{~m}\) long carries a current of \(13.0 \mathrm{~A}\) and makes an angle of \(35.0^{\circ}\) with a uniform magnetic field of magnitude \(\underline{B}=\) \(1.50 \mathrm{~T}\). Calculate the magnetic force on the wire.

An electron is accelerated from rest by a potential difference of \(350 \mathrm{~V}\). It then enters a uniform magnetic field of magnitude \(200 \mathrm{mT}\) with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

A \(13.0 \mathrm{~g}\) wire of length \(L=62.0 \mathrm{~cm}\) is suspended by a pair of flexible leads in a uniform magnetic field of magnitude \(0.440 \mathrm{~T}\) (Fig. \(28-41\) ). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?

A wire \(50.0 \mathrm{~cm}\) long carries a \(0.500 \mathrm{~A}\) current in the positive direction of an \(x\) axis through a magnetic field \(\vec{B}=\) \((3.00 \mathrm{mT}) \hat{\mathrm{j}}+(10.0 \mathrm{mT}) \mathrm{k}\). In unit-vector notation, what is the magnetic force on the wire?

A circular wire loop of radius \(15.0 \mathrm{~cm}\) carries a current of \(2.60 \mathrm{~A}\). It is placed so that the normal to its plane makes an angle of \(41.0^{\circ}\) with a uniform magnetic field of magni- tude \(12.0\) T. (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop?

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