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An electron follows a helical path in a uniform magnetic field given by \(\vec{B}=(20 \hat{\mathrm{i}}-50 \hat{\mathrm{j}}-30 \hat{\mathrm{k}}) \mathrm{mT}\). At time \(t=0\), the electron's velocity is given by \(\vec{v}=(40 \hat{\hat{i}}-30 \hat{j}+50 \hat{k}) \mathrm{m} / \mathrm{s}\). (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) \(\phi = 86.68\degree\); (b) speed is constant; (c) \(\phi\) is constant; (d) \(r = 1.17 \times 10^{-3}\) m.

Step by step solution

01

Understanding the angle between vectors

To find the angle \(\phi\) between the velocity \(\vec{v}\) and magnetic field \(\vec{B}\), use the dot product formula: \(\vec{v} \cdot \vec{B} = |\vec{v}| |\vec{B}| \cos(\phi)\). We will compute the magnitude of \(\vec{v}\) and \(\vec{B}\), and their dot product.
02

Calculate magnitudes of vectors

First, calculate each vector's magnitude: \(|\vec{v}| = \sqrt{40^{2} + (-30)^{2} + 50^{2}} = \sqrt{1600 + 900 + 2500} = \sqrt{5000} = 70.71 \, \mathrm{m/s}\).\(|\vec{B}| = \sqrt{(20 \times 10^{-3})^{2} + (-50 \times 10^{-3})^{2} + (-30 \times 10^{-3})^{2}} = \sqrt{0.004 + 0.025 + 0.009} = \sqrt{0.038} = 0.1949 \, \mathrm{T}\).
03

Compute the dot product

Calculate the dot product: \(\vec{v} \cdot \vec{B} = (40)(20 \times 10^{-3}) + (-30)(-50 \times 10^{-3}) + (50)(-30 \times 10^{-3}) = 0.8 + 1.5 - 1.5 = 0.8\).
04

Determine the angle \(\phi\)

Now solve for \(\cos(\phi)\) using: \(\cos(\phi) = \frac{\vec{v} \cdot \vec{B}}{|\vec{v}| |\vec{B}|} = \frac{0.8}{(70.71)(0.1949)} = \frac{0.8}{13.78} = 0.058\). \(\phi = \cos^{-1}(0.058) = 86.68\degree\).
05

Analyze changes in speed (b)

The speed of an electron in a magnetic field remains constant perpendicular to the field. Hence, speed does not change with time.
06

Analyze changes in angle \(\phi\) (c)

Since the velocity is perpendicular to the magnetic field in a helical path, the angle \(\phi\) remains constant as long as the magnetic field and velocity magnitude remain constant.
07

Calculate radius of helical path (d)

The radius \(r\) of the helical path is given by \(r = \frac{m v_{\perp}}{qB}\), where \(v_{\perp} = v \sin(\phi)\). Using \(v_{\perp} = 40 \sin(86.68\degree) = 39.92 \mathrm{m/s}\), and \( q = 1.6 \times 10^{-19} \mathrm{C}\), and \( B = 0.1949 \mathrm{T}\), \( r = \frac{(9.11 \times 10^{-31} \mathrm{kg})(39.92)}{(1.6 \times 10^{-19} \mathrm{C})(0.1949)} = 1.17 \times 10^{-3} \mathrm{m} \).

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Key Concepts

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

Helical Motion
In physics, helical motion describes the spiral trajectory of charged particles moving through a uniform magnetic field. This rounded path results from a combination of circular and linear motions. When a charged particle, such as an electron, enters a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field direction. This causes the particle to follow a corkscrew-like path.
A key aspect of helical motion is the radius of the helix, which is determined by the perpendicular component of the particle's velocity to the magnetic field. The pitch of the helix, or the distance between the coils, is dictated by the component of the velocity that is parallel to the field. Both the radius and pitch depend on the initial velocity and the magnetic field's strength. Understanding these components helps predict and describe the behavior of charged particles in fields, which is fundamental in applications like cyclotrons and magnetic resonance imaging (MRI).
Lorentz Force
The Lorentz force is the fundamental electromagnetic force acting on a charged particle moving through an electric or magnetic field. For a magnetic field, this force is given by the equation: \[\vec{F} = q(\vec{v} \times \vec{B})\]where \(q\) is the charge of the particle, \(\vec{v}\) is its velocity, and \(\vec{B}\) is the magnetic field. The force is always perpendicular to both the velocity and the magnetic field, which is why it results in circular or helical motion rather than accelerating the particle.
Since the magnetic force does not do work on the particle (as it acts perpendicular to the direction of motion), the speed of the particle remains constant. Only the direction of motion changes, which leads to continuous circular or spiral motion in case of magnetic fields, forming the basis of the helical path described in our problem.
Velocity Vectors
Velocity vectors are a fundamental concept in mechanics, representing both the speed and direction of an object. In the context of this exercise, the initial velocity vector \( \vec{v} = (40 \hat{i} - 30 \hat{j} + 50 \hat{k}) \mathrm{m/sec} \) describes the electron's initial speed and trajectory through 3D space. Each component of the vector corresponds to the motion in the \(x\), \(y\), and \(z\) directions, respectively.
Understanding the velocity vector is crucial as it helps determine the path followed by the electron in a magnetic field. The vector's direction indicates the initial path, while its magnitude, or the speed, affects how the electron interacts with the magnetic field. In combination with the magnetic field vector, it guides us in determining motion characteristics, such as the pitch and radius of the helical path.
Dot Product
The dot product, also known as scalar product, is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. In our problem, it is used to determine the angle \( \phi \) between two vectors: the velocity \( \vec{v} \) and the magnetic field \( \vec{B} \). The formula for the dot product is:\[\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\phi)\]This operation is especially useful as it offers a method to quantify the directional alignment between vectors. If the result is zero, they are perpendicular; a positive result means they point in a generally same direction, while a negative result indicates opposite directions.
In this exercise, finding the dot product of \( \vec{v} \) and \( \vec{B} \), and solving for \( \phi \), gives critical insight into the orientation of the electron's path with respect to the magnetic field. This orientation is essential to understanding and calculating the characteristics of the helical motion the electron undergoes in the magnetic field.

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Most popular questions from this chapter

Figure 28-22 shows a metallic block, with its faces parallel to coordinate axes. The block is in a uniform magnetic field of magnitude \(0.020 \mathrm{~T}\). One edge length of the block is \(32 \mathrm{~cm}\); the block is not drawn to scale. The block is moved at \(3.5 \mathrm{~m} / \mathrm{s}\) parallel to each axis, in turn, and the resulting potential difference \(V\) that appears across the block is measured. With the motion parallel to the \(y\) axis, \(V=12 \mathrm{mV}\); with the motion parallel to the \(z\) axis, \(V=18 \mathrm{mV}\); with the motion parallel to the \(x\) axis, \(V=0\). What are the block lengths (a) \(d_{x}\), (b) \(d_{y}\), and (c) \(d_{z} ?\)

(a) What uniform magnetic field, applied perpendicular to a beam of electrons moving at \(1.30 \times 10^{6} \mathrm{~m} / \mathrm{s}\), is required to make the electrons travel in a circular arc of radius \(0.500 \mathrm{~m} ?(\mathrm{~b})\) What is the period of the motion?

A source injects an electron of speed \(v=1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) 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.

An ion source is producing \({ }^{6} \mathrm{Li}\) ions, which have charge \(+e\) and mass \(9.99 \times 10^{-27} \mathrm{~kg}\). The ions are accelerated by a potential difference of \(25 \mathrm{kV}\) and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude \(B=1.2 \mathrm{~T}\). Calculate the strength of the electric field, to be set up over the same region, that will allow the \({ }^{6} \mathrm{Li}\) ions to pass through without any deflection.

An electron follows a helical path in a uniform magnetic field of magnitude \(1.30 \mathrm{~T}\). The pitch of the path is \(6.00 \mu \mathrm{m}\), and the magnitude of the magnetic force on the electron is \(2.00 \times 10^{-14} \mathrm{~N}\). What is the electron's speed?

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