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Chapter 28: Problem 34

An electron follows a helical path in a uniform magnetic field given by \(\vec{B}=(20 \hat{i}-50 \hat{j}-30 \hat{k}) \mathrm{mT}\). At time \(t=0\), the electron's velocity is given by \(\vec{v}=(20 \hat{\mathrm{i}}-30 \hat{\mathrm{j}}+50 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s}\). (a) What is the angle \(\phi\) between \(\vec{v}\) and \(\vec{B} ?\) The electron's velocity changes with time. Do (b) its speed and (c) the angle \(\phi\) change with time? (d) What is the radius of the helical path?

Short Answer

Expert verified
(a) 84.06掳, (b) Speed constant, (c) Angle constant, (d) Radius = 5.7 mm.

Step by step solution

01

Find the angle between velocity and magnetic field

To find the angle \( \phi \) between \( \vec{v} \) and \( \vec{B} \), we use the dot product formula: \( \vec{v} \cdot \vec{B} = |\vec{v}| |\vec{B}| \cos(\phi) \). First, calculate \( \vec{v} \cdot \vec{B} = (20)(20) + (-30)(-50) + (50)(-30) = 400 + 1500 - 1500 = 400 \). Next, find the magnitudes: \( |\vec{v}| = \sqrt{20^2 + (-30)^2 + 50^2} = \sqrt{400 + 900 + 2500} = \sqrt{3800} = 61.64 \) m/s and \( |\vec{B}| = \sqrt{20^2 + (-50)^2 + (-30)^2} = \sqrt{400 + 2500 + 900} = \sqrt{3800} = 61.64 \) mT. Finally, solve for \( \phi \): \( \cos(\phi) = \frac{400}{61.64 \times 61.64} \approx 0.1058 \), and thus, \( \phi \approx \cos^{-1}(0.1058) \approx 84.06^\circ \).
02

Analyze change in speed and angle

In a uniform magnetic field, an electron's speed does not change because the magnetic force is always perpendicular to the velocity, providing centripetal force with no work done. Hence, the speed remains constant. The angle \( \phi \) can fluctuate over time as it represents the alignment of instantaneous velocity and magnetic field. However, in uniform magnetic fields with a constant electron energy, \( \phi \) also remains constant.
03

Calculate the radius of the helical path

The radius of the helical path \( r \) is determined by the component of the velocity perpendicular to the magnetic field, \( v_{\perp} = v \sin(\phi) \). Given \( v = 61.64 \) m/s and \( \phi \approx 84.06^\circ \), \( v_{\perp} = 61.64 \times \sin(84.06^\circ) \approx 61.33 \) m/s. The radius \( r \) is found using \( r = \frac{m v_{\perp}}{q B} \), where \( m \) is the mass of the electron, \( q \) is the charge of the electron, and \( B = 61.64 \times 10^{-3} \) T. Substituting the known values, \( r = \frac{9.11 \times 10^{-31} \times 61.33}{1.6 \times 10^{-19} \times 61.64 \times 10^{-3}} \approx 5.7 \times 10^{-3} \) m.

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Key Concepts

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

Electron Motion
When an electron moves through a magnetic field, it experiences a force due to its charge. This force acts perpendicular to both the direction of its velocity and the magnetic field. This unique orientation causes the electron to follow a circular path, rather than moving in a straight line.
This path can appear as a helix if there is an additional component of motion parallel to the magnetic field. Understanding the electron's motion requires analyzing both its speed and direction. While the magnetic field does not change the speed of the electron because no work is done, it alters the trajectory, affecting how the electron moves through space.
Helical Path
An electron in a magnetic field doesn't just spin in circles. If there is a velocity component along the magnetic field, the electron traces a helical or spiral path. This happens because:
  • The field affects the perpendicular component of the electron's velocity, maintaining a circular motion for that part.
  • The parallel component remains unaffected, allowing the electron to advance along the field lines, creating the helix.
Thus, the path results from the combination of circle and line. This helix is defined by two main parameters: the radius determined by the perpendicular velocity component and the pitch, which is the distance it advances along the field in one complete circle.
Angle Calculation
To determine the angle between the electron's velocity and the magnetic field, the dot product formula is used. This method calculates the cosine of the angle based on the known vectors of velocity and magnetic field. Here, the steps are:
  • Calculate the dot product \( \vec{v} \cdot \vec{B} \).
  • Determine the magnitudes \( |\vec{v}| \) and \( |\vec{B}| \).
  • Use the relationship \( \cos(\phi) = \frac{\vec{v} \cdot \vec{B}}{|\vec{v}| |\vec{B}|} \).
The calculated angle \( \phi \) reflects how the velocity vector aligns with the magnetic field. Understanding \( \phi \) is crucial, as it demonstrates the impact of the magnetic field on the electron's trajectory.
Velocity Components
The velocity of an electron moving in a magnetic field can be broken into two components, as it's crucial for understanding its motion:
  • Perpendicular Component: This is the portion of velocity at right angles to the magnetic field. It determines the radius of the circular motion.
  • Parallel Component: This is along the magnetic field line. It remains unaffected, allowing forward motion along the field.
By combining these components, we can fully describe the electron's motion. Calculating the perpendicular velocity involves the angle \( \phi \), showing that even though the speed doesn't change, breaking it down helps understand the motion dynamics. Ultimately, the particle's velocity components define the characteristics of its path, illustrating the influence of the magnetic field.

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Most popular questions from this chapter

A wire lying along a \(y\) axis from \(y=0\) to \(y=0.250 \mathrm{~m}\) carries a current of \(2.00 \mathrm{~m} \mathrm{~A}\) in the negative direction of the axis. The wire fully lies in a nonuniform magnetic field that is given by \(\vec{B}=(0.300 \mathrm{~T} / \mathrm{m}) y \hat{\mathrm{i}}+(0.400 \mathrm{~T} / \mathrm{m}) y \mathrm{j} .\) In unit-vector notation, what is the magnetic force on the wire?

An electron has an initial velocity of \((12.0 \hat{\mathrm{j}}+15.0 \mathrm{k}) \mathrm{km} / \mathrm{s}\) and a constant acceleration of \(\left(2.00 \times 10^{12} \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}\) in a region in which uniform electric and magnetic fields are present. If \(\vec{B}=(400 \mu \mathrm{T}) \hat{\mathrm{i}}\), find the electric field \(\vec{E}\).

In a nuclear experiment a proton with kinetic energy \(1.0 \mathrm{MeV}\) moves in a circular path in a uniform magnetic field. What energy must (a) an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) and (b) a deuteron \((q=+e, m=2.0 \mathrm{u})\) have if they are to circulate in the same circular path?

An electron moves in a circle of radius \(r=5.29 \times 10^{-11} \mathrm{~m}\) with speed \(2.19 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Treat the circular path as a current loop with a constant current equal to the ratio of the electron's charge magnitude to the period of the motion. If the circle lies in a uniform mag. netic field of magnitude \(B=\) \(7.10 \mathrm{mT}\), what is the maximum possible magnitude of the torque produced on the loop by the field?

A strip of copper \(150 \mu \mathrm{m}\) thick and \(4.5 \mathrm{~mm}\) wide is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(0.65 \mathrm{~T}\), with \(\vec{B}\) perpendicular to the strip. A current \(i=23 \mathrm{~A}\) is then sent through the strip such that a Hall potential difference \(V\) appears across the width of the strip. Calculate \(V\). (The number of charge carriers per unit volume for copper is \(8.47 \times\) \(10^{28}\) electrons/m \(^{3}\).)
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