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Chapter 10: Problem 9

In a magnetic field, field lines are curves to which the magnetic induction \(\mathbf{B}\) is everywhere tangential. By evaluating \(d \mathbf{B} / d s\), where \(s\) is the distance measured along a field line, prove that the radius of curvature at any point on a line is given by $$ \rho=\frac{B^{3}}{|\mathbf{B} \times(\mathbf{B} \cdot \nabla) \mathbf{B}|} $$

Short Answer

Expert verified
Radius of curvature, \( \rho \), for magnetic field lines is \( \frac{B^3}{|\mathbf{B} \times (\mathbf{B} \cdot abla) \mathbf{B}|} \).

Step by step solution

01

Understand the Problem

You need to prove that the radius of curvature (\( \rho \)) of magnetic field lines is given by the formula \( \rho = \frac{B^3}{|\mathbf{B} \times (\mathbf{B} \cdot abla) \mathbf{B}|} \). To do this, start by evaluating the derivative of \( \mathbf{B} \) with respect to \( s \), the distance along the field line.
02

Define Tangential Relationship

Since \( \mathbf{B} \) is tangential to the field lines, write \( \mathbf{T} = \mathbf{B}/B \) where \( \mathbf{T} \) is the unit tangent vector along the field line. Therefore, \( \frac{d \mathbf{B}}{d s} \) must include the derivative of \( \mathbf{T} \).
03

Use Chain Rule

Apply the chain rule: \( \frac{d \mathbf{B}}{d s} = \frac{d}{d s}(B \mathbf{T}) = \frac{d B}{d s} \cdot \mathbf{T} + B \cdot \frac{d \mathbf{T}}{d s} \). Note that \( \frac{d B}{d s} \) is the directional derivative of \( B \) along the field line.
04

Relate to Curvature

Recall that the radius of curvature \( \rho \) is given by \( \frac{1}{k} \) where \( k \) is the curvature. The curvature is defined by \( k = \left| \frac{d \mathbf{T}}{d s} \right| \). Thus, we need to find \( \frac{d \mathbf{T}}{d s} \).
05

Express the Unit Tangent Derivative

Find \( \frac{d \mathbf{T}}{d s} \) by using \( \mathbf{T} \): \( \frac{d \mathbf{T}}{d s} = \frac{d}{d s} \left( \frac{\mathbf{B}}{B} \right) = \frac{ B \frac{d \mathbf{B}}{d s} - \mathbf{B} \frac{d B}{d s}}{B^2} = \frac{1}{B} \left( \frac{d \mathbf{B}}{d s} - \mathbf{T} \frac{d B}{d s} \right) \).
06

Use Gradient Term

Recognize that \( \frac{d B}{d s} = \mathbf{T} \cdot abla B \), so \( \frac{d \mathbf{T}}{d s} = \frac{1}{B} \left( \frac{d \mathbf{B}}{d s} - \mathbf{T} (\mathbf{T} \cdot abla B) \right) \). This simplifies using the definition of the directional derivative.
07

Calculate Field Derivative using Vector Operations

From above equations, \( \frac{d \mathbf{B}}{d s} = (\mathbf{T} \cdot abla) \mathbf{B} = \frac{\mathbf{B} \cdot abla \mathbf{B}}{B} \), rephrase it to help extract the curvature formula. Then use \( \mathbf{B} \cdot abla \mathbf{B} \) cross product property.
08

Final Expression for Radius

Combine all terms and show that the formula for curvature radius \( \rho \) is indeed \( \rho = \frac{B^3}{|\mathbf{B} \times (\mathbf{B} \cdot abla) \mathbf{B}|} \). This requires evaluating magnitude and proper cross product properties for vector fields.

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Key Concepts

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

Radius of Curvature
The radius of curvature, often denoted as \( \rho \), is a measure of the bending of a curve at a particular point. For magnetic field lines, it tells us how sharply the lines are curved. The formula we aim to prove in the context of magnetic induction is:
\( \rho = \frac{B^3}{|\textbf{B} \times (\textbf{B} \boldsymbol{abla}) \textbf{B}|} \).

This equation means that the radius of curvature is directly proportional to the cube of the magnetic field's strength \( B \) and inversely proportional to the magnitude of the vector product of the magnetic field \( \textbf{B} \) and the result of its directional derivative through space.
To approach this proof, we need to understand how magnetic field lines relate to the tangent vector \( \textbf{T} \) and use vector calculus to manipulate these components.

Understanding curvature helps in various fields such as electromagnetism and mechanics, where forces and motions are impacted by the bending of field lines. In this specific case, this insight provides an understanding of how concentrated the magnetic force is along the curve of the lines.
Vector Calculus
Vector calculus plays a huge role in the analysis of magnetic field lines. When we say the magnetic induction \( \textbf{B} \) is tangential to the field lines, it implies that at any point, the direction of \( \textbf{B} \) is the same as that of the tangent vector to the line.

To proceed with the solution, we utilize key principles such as:
  • Directional Derivative: Measures the rate of change of the function (in this case, \( B \)) in the direction of a vector (the unit tangent vector \( \textbf{T} \)). The notation used is \( \textbf{T} \boldsymbol{abla} B \).
  • Unit Tangent Vector: Given by \( \textbf{T} = \frac{\textbf{B}}{B} \), it is crucial in describing the direction along the field line.
  • Cross Product: It's a binary operation on two vectors in three-dimensional space, symbolized by \( \times \) and used to find perpendicular vectors to the plane containing the original vectors.
The interaction among these concepts helps simplify complex expressions and is key to arriving at the formula for the radius of curvature.
Magnetic Induction
Magnetic induction, indicated by the vector \( \textbf{B} \), refers to the magnetic field created by a current or changing electric field.
In the context of magnetic field lines, the field \( \textbf{B} \) governs the strength and direction at each point. This means the magnetic force felt is along these curved lines which are smooth paths showing the direction of the magnetic field.

To establish the radius of curvature, it’s necessary to comprehend how the magnetic field varies along these paths. This insight was core to deriving the expression below:
\( \frac{d \textbf{B}}{ds} = (\textbf{T} \boldsymbol{abla}) \textbf{B} \).
This is known as the derivative of \( \textbf{B} \) with respect to \( s \), the distance along the field line.

Subsequently, we notice that the change in \( \textbf{B} \)’s direction relative to this varying path directly ties into the field's strength and curvature. Understanding this relationship between the magnetic induction and geometric curvature helps visualize and analyze magnetic phenomena, enhancing our grasp of electromagnetism concepts otherwise abstract.

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Most popular questions from this chapter

At time \(t=0\), the vectors \(\mathbf{E}\) and \(\mathbf{B}\) are given by \(\mathbf{E}=\mathbf{E}_{0}\) and \(\mathbf{B}=\mathbf{B}_{0}\), where the fixed unit vectors \(\mathbf{E}_{0}\) and \(\mathbf{B}_{0}\) are orthogonal. The equations of motion are $$ \begin{aligned} &\frac{d \mathbf{E}}{d t}=\mathbf{E}_{0}+\mathbf{B} \times \mathbf{E}_{0} \\ &\frac{d \mathbf{B}}{d t}=\mathbf{B}_{0}+\mathbf{E} \times \mathbf{B}_{0} \end{aligned} $$ Find \(\mathbf{E}\) and \(\mathbf{B}\) at a general time \(t\), showing that after a long time the directions of \(\mathbf{E}\) and \(\mathbf{B}\) have almost interchanged.

(a) Parameterising the hyperboloid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 $$ by \(x=a \cos \theta \sec \phi, y=b \sin \theta \sec \phi, z=c \tan \phi\), show that an area element on its surface is $$ d S=\sec ^{2} \phi\left[c^{2} \sec ^{2} \phi\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+a^{2} b^{2} \tan ^{2} \phi\right]^{1 / 2} d \theta d \phi $$ (b) Use this formula to show that the area of the curved surface \(x^{2}+y^{2}-z^{2}=a^{2}\) between the planes \(z=0\) and \(z=2 a\) is $$ \pi a^{2}\left(6+\frac{1}{\sqrt{2}} \sinh ^{-1} 2 \sqrt{2}\right) $$

(a) Simplify $$ \nabla \times \mathbf{a}(\nabla \cdot \mathbf{a})+\mathbf{a} \times[\nabla \times(\nabla \times \mathbf{a})]+\mathbf{a} \times \nabla^{2} \mathbf{a} $$ (b) By explicitly writing out the terms in Cartesian coordinates prove that $$ [\mathbf{c} \cdot(\mathbf{b} \cdot \nabla)-\mathbf{b} \cdot(\mathbf{c} \cdot \nabla)] \mathbf{a}=(\nabla \times \mathbf{a}) \cdot(\mathbf{b} \times \mathbf{c}) $$ (c) Prove that \(\mathbf{a} \times(\nabla \times \mathbf{a})=\nabla\left(\frac{1}{2} a^{2}\right)-(\mathbf{a} \cdot \nabla) \mathbf{a}\).

The general equation of motion of a (non-relativistic) particle of mass \(m\) and charge \(q\) when it is placed in a region where there is a magnetic field \(\mathbf{B}\) and an electric field \(\mathbf{E}\) is $$ m \ddot{\mathbf{r}}=q(\mathbf{E}+\dot{\mathbf{r}} \times \mathbf{B}) $$ here \(\mathbf{r}\) is the position of the particle at time \(t\) and \(\dot{\mathbf{r}}=d \mathbf{r} / d t\) etc. Write this as three separate equations in terms of the Cartesian components of the vectors involved. For the simple case of crossed uniform fields \(\mathbf{E}=E \mathbf{i}, \mathbf{B}=B \mathbf{j}\) in which the particle starts from the origin at \(t=0\) with \(\dot{\mathbf{r}}=v_{0} \mathbf{k}\), find the equations of motion and show the following: (a) if \(v_{0}=E / B\) then the particle continues its initial motion; (b) if \(v_{0}=0\) then the particle follows the space curve given in terms of the parameter \(\xi\) by $$ x=\frac{m E}{B^{2} q}(1-\cos \xi), \quad y=0, \quad z=\frac{m E}{B^{2} q}(\xi-\sin \xi) $$ Interpret this curve geometrically and relate \(\xi\) to \(t\). Show that the total distance travelled by the particle after time \(t\) is $$ \frac{2 E}{B} \int_{0}^{t}\left|\sin \frac{B q t^{\prime}}{2 m}\right| d t^{\prime} $$

For the twisted space curve \(y^{3}+27 a x z-81 a^{2} y=0\), given parametrically by $$ x=a u\left(3-u^{2}\right), \quad y=3 a u^{2}, \quad z=a u\left(3+u^{2}\right) $$ show the following: (a) that \(d s / d u=3 \sqrt{2} a\left(1+u^{2}\right)\), where \(s\) is the distance along the curve measured from the origin; (b) that the length of the curve from the origin to the Cartesian point \((2 a, 3 a, 4 a)\) is \(4 \sqrt{2 a}\) (c) that the radius of curvature at the point with parameter \(u\) is \(3 a\left(1+u^{2}\right)^{2}\); (d) that the torsion \(\tau\) and curvature \(\kappa\) at a general point are equal; (e) that any of the Frenet-Serret formulae that you have not already used directly are satisfied.
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