Problem 34 - Fundamentals of Physics

Question

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 elec- tron's velocity is given by \(\vec{v}=(20 \hat{\mathrm{i}}-30 \mathrm{j}+50 \hat{\mathrm{k}}) \mathrm{m} / \mathrm{s} .\) (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?

Step-by-Step Solution

Answer

(a) The angle \(\phi\) is \(\cos^{-1}(1/9.5)\). (b) Speed does not change. (c) The angle \(\phi\) does not change. (d) Calculate using \(R = \frac{m v_{\perp}}{qB}\).

Step 1: Calculate the Dot Product of Vectors

First, calculate the dot product of the velocity vector \(\vec{v}\) and the magnetic field vector \(\vec{B}\). The formula for the dot product of two vectors \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is \(\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\). Plugging in the given vectors, we get:\[\vec{v} \cdot \vec{B} = (20)(20) + (-30)(-50) + (50)(-30) = 400 + 1500 - 1500 = 400\]

Step 2: Calculate the Magnitudes of Vectors

Next, calculate the magnitudes of \(\vec{v}\) and \(\vec{B}\). The magnitude of a vector \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) is given by \(|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\).For \(\vec{v}\) and \(\vec{B}\), we have:\[|\vec{v}| = \sqrt{20^2 + (-30)^2 + 50^2} = \sqrt{400 + 900 + 2500} = \sqrt{3800}\]\[|\vec{B}| = \sqrt{20^2 + (-50)^2 + (-30)^2} = \sqrt{400 + 2500 + 900} = \sqrt{3800}\]

Step 3: Find Angle Using Dot Product

Now, find the angle \(\phi\) using the relation between dot product and cosine of the angle:\[\vec{v} \cdot \vec{B} = |\vec{v}||\vec{B}|\cos(\phi)\]Plugging in the values, we have:\[400 = \sqrt{3800} \cdot \sqrt{3800} \cdot \cos(\phi)\]\[ 400 = 3800 \cdot \cos(\phi) \]Solving for \(\cos(\phi)\), we get \(\cos(\phi) = \frac{400}{3800} = \frac{1}{9.5}\).Find \(\phi\) using \(\phi = \cos^{-1}(\frac{1}{9.5})\).

Step 4: Analyze Change in Speed

(b) The speed of an electron in a uniform magnetic field remains constant because the magnetic force is always perpendicular to the velocity and does not do work. Therefore, the speed will not change over time.

Step 5: Analyze Change in Angle

(c) The angle \(\phi\) will also remain constant over time in a uniform magnetic field because there are no additional forces acting on the electron to change the direction of its velocity vector relative to the magnetic field. The motion is purely rotational.

Step 6: Calculate Radius of Helical Path

(d) The radius \(R\) of the helical path can be found using the formula for the radius of the circular motion of a charged particle in a magnetic field: \[R = \frac{m v_{\perp}}{qB}\]where \(v_{\perp}\) is the component of the velocity perpendicular to the magnetic field, \(m\) is the mass of the electron, and \(q\) is the charge of an electron.First, calculate \(v_{\perp}\):\[v_{\perp} = |\vec{v}| \sin(\phi) = \sqrt{3800} \cdot \sin(\phi)\]Using \(\sin(\phi) = \sqrt{1 - \cos^2(\phi)} = \sqrt{1 - (\frac{1}{9.5})^2}\), calculate \(v_{\perp}\). Then, evaluate \(R\) knowing \[m = 9.11 \times 10^{-31} kg,\ q = 1.60 \times 10^{-19} C,\ |\vec{B}| = 20 \times 10^{-3} T\].

Key Concepts

Helical Path

The motion of an electron in a uniform magnetic field is quite fascinating. When an electron moves with a velocity that isn't aligned with the magnetic field, it doesn't just move in a straight line. Instead, it describes a helical path. This helical motion is a combination of two different types of movement:

- Circular motion: The electron circles around the direction of the magnetic field. - Linear motion: Simultaneously, it travels along the field lines.
This results in a spiral-like path that can be compared to how a spring is coiled. The forces involved ensure that only the component of velocity parallel to the magnetic field leads to forward motion, while the component perpendicular results in circular motion.

Angle Between Vectors

The angle electrons traverse between their velocity and the magnetic field is crucial for determining the nature of their path. This angle, known as \( \phi \), can be calculated using the dot product of the two vectors. The dot product relates to the cosine of the angle between vectors. By calculating \( \cos(\phi) \), you determine how much of the electron's velocity allows for circular motion versus linear motion. Importantly, the value of \( \phi \) doesn't change over time in a uniform magnetic field. As there's no external force altering the electron's course, \( \phi \) remains as initially defined, maintaining a steady path.

Perpendicular Velocity

When we talk about perpendicular velocity, \( v_{\perp} \), we are referring to the component of the electron's velocity that is perpendicular to the magnetic field. This is the component that causes the electron to spiral around the field lines, thus creating the circular part of the helical path.

The perpendicular velocity can be found by using the sine of the angle \( \phi \) between the velocity vector and the magnetic field vector. Specifically, \( v_{\perp} = |\vec{v}| \sin(\phi) \). This component is crucial because it plays a significant role in determining the radius of the circle formed as part of the helical trajectory.

It's fascinating to note that although the velocity of the electron is changing direction, its magnitude, or speed, remains constant. This is because the magnetic force, while constantly redirecting the electron, does not work on it.

Circular Motion Radius

The radius of the circular motion part of the helical path depends on several factors and is important in defining the path's size and nature. It is defined by the formula \( R = \frac{mv_{\perp}}{qB} \), where:

• \( m \) is the mass of the electron, which is a known constant.
• \( v_{\perp} \) is the perpendicular component of the electron's velocity.
• \( q \) is the elementary charge of the electron.
• \( B \) is the magnitude of the magnetic field.

The radius tells you how tightly the electron spirals. A larger perpendicular velocity or smaller magnetic field would result in a larger radius. Likewise, a smaller mass would see a larger radius, affecting the electron's helical path diameter. Understanding this radius is key to predicting the electron's behavior in different magnetic fields.