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Chapter 29: Problem 12

Source of Electrons A source injects an electron of speed \(v=\) \(1.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\) into a uniform magnetic field of magnitude \(B=\) \(1.0 \times 10^{-3} \mathrm{~T}\). The velocity of the electron makes an angle \(\theta=10^{\circ}\) with the direction of the magnetic field. Find the distance \(d\) from the point of injection at which the electron next crosses the field line that passes through the injection point.

Short Answer

Expert verified
The distance is approximately 0.5328 meters.

Step by step solution

01

- Determine the Centripetal Force

Firstly, recall the formula for the force exerted on a moving charge in a magnetic field, which is given by \[ F = qvB \sin(\theta) \] where, for an electron, the charge \( q = -1.6 \times 10^{-19} \mathrm{C} \), the velocity \( v = 1.5 \times 10^{7} \mathrm{~m} / \, \mathrm{s}\), the magnetic field strength \( B = 1.0 \times 10^{-3} \mathrm{T} \), and the angle \( \theta = 10^{\circ} \). The magnitude of the force becomes \[ |F| = |q| v B \sin(\theta) \].
02

- Calculate the Centripetal Force

Substitute the given values into the equation from Step 1:\[ |F| = (1.6 \times 10^{-19} \mathrm{C})(1.5 \times 10^{7} \mathrm{m/s})(1.0 \times 10^{-3} \mathrm{T}) \sin(10^{\circ}) \]Using \( \sin(10^{\circ}) \approx 0.1736 \)\[ |F| \approx (1.6 \times 10^{-19})(1.5 \times 10^{7})(1.0 \times 10^{-3})(0.1736) \approx 4.16 \times 10^{-15} \mathrm{N} \]
03

- Determine the Cyclotron Frequency

The electron will move in helical motion due to the magnetic field. To find the distance \( d \), start by finding the cyclotron frequency (frequency of circular motion):\[ f = \frac{|q| B}{2 \pi m} \]where mass of the electron \( m = 9.11 \times 10^{-31} \mathrm{kg} \). Substituting the values:\[ f = \frac{(1.6 \times 10^{-19})(1.0 \times 10^{-3})}{2 \pi (9.11 \times 10^{-31})} \approx 2.78 \times 10^{7} \mathrm{Hz} \]
04

- Calculate the Time Period of Circular Motion

The time period \( T \) of the circular motion can be calculated as the inverse of the frequency:\[ T = \frac{1}{f} \approx \frac{1}{2.78 \times 10^{7}} \approx 3.6 \times 10^{-8} \mathrm{seconds} \]
05

- Calculate the Pitch of the Helical Path

The pitch of the helical path (distance between successive crossings of the magnetic field line) is given by:\[ d = v_{||} T \] where \(v_{||}\) is the component of velocity parallel to the magnetic field:\[ v_{||} = v \cos(10^{\circ}) \approx (1.5 \times 10^{7}) \cos(10^{\circ}) \approx 1.48 \times 10^{7} \mathrm{m/s} \]Then,\[ d = (1.48 \times 10^{7})(3.6 \times 10^{-8}) \approx 0.5328 \mathrm{m} \]

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Key Concepts

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

Centripetal Force
In the presence of a magnetic field, a charged particle such as an electron experiences a force that continually changes its direction, making it move in a circular path. This force is called the centripetal force. The formula to calculate the force on a moving charge in a magnetic field is given by . For an electron, the charge , the velocity , the magnetic field strength , and the angle . The magnitude of the force thus becomes: . By solving this equation with the given values and , we get: . This force is what keeps the electron in motion along a circular trajectory within the magnetic field.
Cyclotron Frequency
When an electron moves through a magnetic field, it follows a circular or helical path depending on its angle of entry relative to the field. The frequency at which it revolves around the magnetic field lines is called the cyclotron frequency. It's calculated using , where is the charge of the electron, is the magnetic field strength, and is the mass of the electron. Substituting the given values into the equation: . This means the electron completes approximately cycles per second in the magnetic field. This frequency is crucial to understanding how fast the electron revolves and how its motion evolves over time.
Helical Motion
Due to the magnetic field and the angle at which the electron is injected, its motion follows a helical path. This is because the velocity of the electron has two components: one perpendicular to the magnetic field, which causes circular motion, and one parallel to the magnetic field, which propels the electron along the field lines. This combination of motions results in a helical trajectory. To find the pitch, or the distance between successive crossings of the magnetic field line, we use the formula , where is the component of velocity parallel to the magnetic field and is the time period of the circular motion. The parallel component of the velocity is . Thus, the pitch . This distance indicates how far the electron travels along the field direction before completing one full circle.

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Most popular questions from this chapter

Arbitrarily Shaped Coil Prove that the relation \(\tau=N|i| A B \sin \phi\) (Eq. 29-31) holds for closed loops of arbitary shape and not only for rectangular loops as in Fig. 29-22. (Hint: Replace the loop of arbitrary shape with an assembly of adjacent long, thin, approximately rectangular loops that are nearly equivalent to the loop of arbitrary shape as far as the distribution of current is concerned.)

Force on Charges An electron that has velocity $$ \vec{v}=\left(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(3.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}} $$ moves through the magnetic field \(\vec{B}=(0.030 \mathrm{~T}) \hat{\mathrm{i}}-(0.15 \mathrm{~T}) \hat{\mathrm{j}}\). (a) Find the force on the electron. (b) Repeat your calculation for a proton having the same velocity.

Three Force Fields We have studied three long-range forces: gravity, electricity, and magnetism. Compare and contrast these three forces giving at least one feature that all three forces have in common, and at least one feature that distinguishes each force.

ABCDEFA Figure \(\quad 29-36\) \(\begin{array}{l}\text { shows } a & \text { current } \text { loop }\end{array}\) \(A B C D E F A\) carrying a current \(i\) \(=5.00 \mathrm{~A}\). The sides of the loop are parallel to the coordinate axes, with \(A B=20.0 \mathrm{~cm}, B C=\) \(30.0 \mathrm{~cm}\), and \(F A=10.0 \mathrm{~cm}\). Calculate the magnitude and direction of the magnetic dipole moment of this loop. (Hint: Imagine equal and opposite currents \(i\) in the line segment \(A D\); then treat the two rectangular loops \(A B C D A\) and \(A D E F A .)\)

TV Camera An electron in a TV camera tube is moving at \(7.20 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a magnetic field of strength \(83.0 \mathrm{mT}\). (a) Without knowing the direction of the field, what can you say about the greatest and least magnitudes of the force acting on the electron due to the field? (b) At one point the electron has an acceleration of magnitude \(4.90 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2} .\) What is the angle between the electron's velocity and the magnetic field?
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