Problem 32 - Classical Mechanics

Question

If you have some experience in electromagnetism, you could do the following problem concerning the curious situation illustrated in Figure \(1.8 .\) The electric and magnetic fields at a point \(\mathbf{r}_{1}\) due to a charge \(q_{2}\) at \(\mathbf{r}_{2}\) moving with constant velocity \(\mathbf{v}_{2}\) (with \(v_{2} \ll c\) ) are \(^{15}\) $$\mathbf{E}\left(\mathbf{r}_{1}\right)=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{2}}{s^{2}} \hat{\mathbf{s}} \quad \text { and } \quad \mathbf{B}\left(\mathbf{r}_{1}\right)=\frac{\mu_{0}}{4 \pi} \frac{q_{2}}{s^{2}} \mathbf{v}_{2} \times \hat{\mathbf{s}}$$ where \(\mathbf{s}=\mathbf{r}_{1}-\mathbf{r}_{2}\) is the vector pointing from \(\mathbf{r}_{2}\) to \(\mathbf{r}_{1}\). (The first of these you should recognize as Coulomb's law.) If \(\mathbf{F}_{12}^{\mathrm{el}}\) and \(\mathbf{F}_{12}^{\text {mag }}\) denote the electric and magnetic forces on a charge \(q_{1}\) at \(\mathbf{r}_{1}\) with velocity \(\mathbf{v}_{1},\) show that \(F_{12}^{\operatorname{mag}} \leq\left(v_{1} v_{2} / c^{2}\right) F_{12}^{\mathrm{el}} .\) This shows that in the non-relativistic domain it is legitimate to ignore the magnetic force between two moving charges.

Step-by-Step Solution

Answer

The magnetic force is negligible compared to the electric force for non-relativistic speeds.

Step 1: Write Down Expressions for Forces

The electric force on charge \(q_1\) at \(\mathbf{r}_1\) is given by Coulomb's law:\[ \mathbf{F}_{12}^{\mathrm{el}} = q_1 \mathbf{E}(\mathbf{r}_1) = \frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_1 q_2}{s^2} \hat{\mathbf{s}} \].The magnetic force is given by the Lorentz force law:\[ \mathbf{F}_{12}^{\text{mag}} = q_1 (\mathbf{v}_1 \times \mathbf{B}(\mathbf{r}_1)) \].Substitute \(\mathbf{B}(\mathbf{r}_1)\) into the expression for \(\mathbf{F}_{12}^{\text{mag}}\):\[ \mathbf{F}_{12}^{\text{mag}} = q_1 \frac{\mu_0}{4 \pi} \frac{q_2}{s^2} (\mathbf{v}_1 \times (\mathbf{v}_2 \times \hat{\mathbf{s}})) \].

Step 2: Simplify Magnetic Force Expression

Use the vector triple product identity: \((\mathbf{a} \times (\mathbf{b} \times \mathbf{c})) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\).Substitute into the magnetic force expression:\[ (\mathbf{v}_1 \cdot \hat{\mathbf{s}}) \mathbf{v}_2 - (\mathbf{v}_1 \cdot \mathbf{v}_2) \hat{\mathbf{s}} \].Thus,\[ \mathbf{F}_{12}^{\text{mag}} = \frac{q_1 q_2 \mu_0}{4 \pi s^2} ((\mathbf{v}_1 \cdot \hat{\mathbf{s}}) \mathbf{v}_2 - (\mathbf{v}_1 \cdot \mathbf{v}_2) \hat{\mathbf{s}}) \].

Step 3: Calculate Magnitude of Magnetic Force

The magnitude of \(\mathbf{F}_{12}^{\text{mag}}\) is the sum of the magnitudes of the component vectors:\[ \left|\mathbf{F}_{12}^{\text{mag}}\right| = \frac{q_1 q_2 \mu_0}{4 \pi s^2} \cdot \left| (\mathbf{v}_1 \cdot \hat{\mathbf{s}}) \mathbf{v}_2 - (\mathbf{v}_1 \cdot \mathbf{v}_2) \hat{\mathbf{s}} \right| \].Apply the inequality \(|\mathbf{a} - \mathbf{b}| \leq |\mathbf{a}| + |\mathbf{b}|\) to the magnitudes:\[ \leq \frac{q_1 q_2 \mu_0}{4 \pi s^2} \left(\left|\mathbf{v}_1 \cdot \hat{\mathbf{s}}\right| \left|\mathbf{v}_2\right| + \left|\mathbf{v}_1 \cdot \mathbf{v}_2\right|\right) \].

Step 4: Connect Terms and Apply Approximation

Using \( \mu_0 \epsilon_0 c^2 = 1 \), substitute \( \mu_0 = \frac{1}{\epsilon_0 c^2} \) into the magnetic force expression:\[ \left|\mathbf{F}_{12}^{\text{mag}}\right| \leq \frac{q_1 q_2}{4 \pi \epsilon_0 s^2} \frac{v_2}{c^2} (v_1 + v_2) \].Notice that the electric force is \( \mathbf{F}_{12}^{\mathrm{el}} = \frac{q_1 q_2}{4 \pi \epsilon_0 s^2} \hat{\mathbf{s}} \):\[ \left|\mathbf{F}_{12}^{\text{mag}}\right| \leq \left(\frac{v_1 v_2}{c^2} + \frac{v_2^2}{c^2}\right) \left|\mathbf{F}_{12}^{\mathrm{el}}\right| \],where \(v_2^2/c^2\) is negligible compared to \(v_1 v_2/c^2\).

Step 5: Conclusion and Result

Because \(v_2 \ll c\), the term \(\frac{v_2^2}{c^2}\) can be ignored, leaving only:\[ \left|\mathbf{F}_{12}^{\text{mag}}\right| \leq \frac{v_1 v_2}{c^2} \left|\mathbf{F}_{12}^{\mathrm{el}}\right| \].This shows the magnetic force is negligible compared to the electric force in the non-relativistic domain.

Key Concepts

Electric Field

The electric field is a fundamental concept in electromagnetism. It refers to the region around a charged particle where an electric force can be felt by another charge. In simple terms, imagine a charged object creating an invisible field around it. If another charge enters this field, it experiences a force that is proportional to its own charge and inversely proportional to the square of the distance from the source charge.

The formula for the electric field \(\mathbf{E}\), as in this exercise, is given by:

• \( \mathbf{E}(\mathbf{r}_{1}) = \frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{2}}{s^{2}} \hat{\mathbf{s}} \)

This formula illustrates how the electric field depends on the charge creating it, the distance from the charge, and the permittivity of free space, \(\epsilon_0\).

Essentially, the electric field is a vector that points away from positive charges and towards negative charges. This aligns with the idea of forces in physics where like charges repel and unlike charges attract, following Coulomb's Law.

Magnetic Force

Magnetic force arises from the motion of electric charges. It is a fundamental force in nature, responsible for actions such as the operation of electric motors and the navigation of compasses.

A moving charge, like a current in a wire, produces a magnetic field around it. If another charge moves within this field, it experiences a magnetic force, not aligned with the electric field, but rather perpendicular to both the velocity of the moving charge and the direction of the field.

For this exercise, the magnetic field \(\mathbf{B}(\mathbf{r}_{1})\) produced by a moving charge \(q_{2}\) is expressed as:

• \( \mathbf{B}(\mathbf{r}_{1}) = \frac{\mu_{0}}{4 \pi} \frac{q_{2}}{s^{2}} \mathbf{v}_{2} \times \hat{\mathbf{s}} \)

Here, \(\mu_{0}\) represents the magnetic constant or permeability of free space. The cross product \( \mathbf{v}_{2} \times \hat{\mathbf{s}} \) indicates the force's direction is perpendicular to the plane formed by \( \mathbf{v}_{2} \) and \( \hat{\mathbf{s}} \). Magnetic forces are unique because they require motion; a static charge does not experience a magnetic force.

Lorentz Force

The Lorentz force equation is crucial in understanding how charged particles move in electric and magnetic fields. It combines both electric and magnetic forces acting on a charge. By understanding this, we gain insight into the motion of electrons in fields, a principle that powers many modern technologies like cathode ray tubes and the function of cyclotrons.

The total force \(\mathbf{F}\) on a moving charge \(q\) in external electric \(\mathbf{E}\) and magnetic fields \(\mathbf{B}\) is given by:

• \(\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})\)

This equation shows the superposition of the electric and magnetic forces. The electric part of the Lorentz force is straightforward, simply \(q\mathbf{E}\). The magnetic part, \(q(\mathbf{v} \times \mathbf{B})\), signifies how motion through the magnetic field affects the charge.

The forces interact in such a way that if a charged particle moves through both fields, its path can curve due to the perpendicular magnetic forces, much like in the example problem where motion at speeds much less than light allows simpler approximations.

Coulomb's Law

Coulomb's Law is as fundamental to electromagnetism as Newton's Law of Gravitation is to gravity. It calculates the electric force between two point charges. This foundational law helps us quantify how charges interact and the resultant forces they exert on each other.

The formula for Coulomb's Law in its common form is:

• \( \mathbf{F}_{12}^{\mathrm{el}} = \frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{q_{1} q_{2}}{s^2} \hat{\mathbf{s}} \)

Here, \(\mathbf{F}_{12}^{\mathrm{el}}\) is the force between charges \(q_{1}\) and \(q_{2}\), \(s\) is the distance separating them, and \(\hat{\mathbf{s}}\) is the unit vector pointing from one charge to the other.

This law reveals that the force is inversely proportional to the square of the distance between charges, meaning it rapidly decreases as they move apart. This inverse-square law also helps explain why electric forces can be strong at close range and almost negligible at great distances.