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Integral of \(A\), flux of \(B\) Show that the line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This result is very similar to Ampère's law, which says that the line integral of the magnetic field \(\mathbf{B}\) around a closed curve \(C\) equals (up to a factor of \(\mu_{0}\) ) the current flux \(I\) through a surface \(S\) bounded by the curve.

Short Answer

Expert verified
The line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This is shown by expressing the magnetic flux as a surface integral of the curl of \(\mathbf{A}\) and then applying Stoke's theorem.

Step by step solution

01

Definition of magnetic flux

The magnetic flux \(\Phi\) through a surface \(S\) is defined as the surface integral of the magnetic field \(\mathbf{B}\) over \(S\). This can be mathematically written as \(\Phi = \int_{S} \mathbf{B} \cdot d\mathbf{S}\). In terms of the vector potential \(\mathbf{A}\), the magnetic field \(\mathbf{B}\) can be expressed as \(\mathbf{B} = \nabla × \mathbf{A}\). Therefore, the magnetic flux can be rewritten as \(\Phi = \int_{S} (\nabla × \mathbf{A}) \cdot d\mathbf{S}\).
02

Application of Stoke's theorem

Stoke's theorem is a fundamental theorem in vector calculus that states the surface integral of the curl of a vector field over a surface \(S\) is equal to the line integral of the vector field around the boundary of \(S\). In mathematical terms, \(\int_{S} (\nabla × \mathbf{A}) \cdot d\mathbf{S} = \oint_{C} \mathbf{A} \cdot d\mathbf{r}\), where \(C\) is the boundary of \(S\) and \(d\mathbf{r}\) is an infinitesimal element along \(C\). Applying this theorem to the magnetic flux gives \(\Phi = \oint_{C} \mathbf{A} \cdot d\mathbf{r}\).
03

Conclusion

Therefore, it can be conclusively stated that the line integral of the vector potential \(\mathbf{A}\) around a closed curve \(C\) equals the magnetic flux \(\Phi\) through a surface \(S\) bounded by the curve. This demonstrates the given proposition.

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Key Concepts

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

Vector Potential
The concept of vector potential plays a pivotal role in electromagnetism, particularly in the context of magnetic fields. A vector potential \textbf{A} is a vector field whose curl, according to Maxwell's equations, gives us the magnetic field \textbf{B}. Mathematically, this relationship is expressed as: \( \mathbf{B} = abla \times \mathbf{A} \).

This might seem a tad abstract at first, but you can think of \textbf{A} as a way to 'encode' magnetic field information. Despite there being an infinite number of vectors \textbf{A} that can satisfy this equation for a given \textbf{B} (due to the gauge freedom), the vector potential simplifies many calculations in electromagnetism and can be key to understanding both the flow of the magnetic field and, as in our exercise, the magnetic flux through a surface.
Stoke's Theorem
Stoke’s theorem bridges the gap between line integrals and surface integrals. It can be considered a higher-dimensional version of the fundamental theorem of calculus that relates a line integral around some curve \textbf{C} to a surface integral over the surface \textbf{S} bounded by \textbf{C}.

The theorem declares that the integral of a curl of a vector field over a surface is equal to the line integral of the vector field along the boundary of the surface. The formula for Stoke's theorem is: \( \int_{S} (abla \times \mathbf{A}) \cdot d\mathbf{S} = \oint_{C} \mathbf{A} \cdot d\mathbf{r} \).

This links the local rotation (or curl) of \textbf{A} within \textbf{S} to its circulation around the edge, effectively 'converting' the intuitive two-dimensional concept of circulation into the more abstract three-dimensional realm.
Line Integral
The line integral is a key concept in calculus and physics, especially when dealing with vector fields. It represents the work done by a force field in moving a particle along a path \textbf{C}. When it comes to the vector potential \textbf{A}, the line integral around a closed curve \textbf{C} (notated as \( \oint_{C} \)) is found by summing the dot product of \textbf{A} with the differential element of the path \textbf{dr} all along \textbf{C}.

As the definition suggests, it's expressed mathematically as \( \oint_{C} \mathbf{A} \cdot d\mathbf{r} \). The calculation of a line integral involves determining how much of the vector field is aligned with the path at all points, then 'adding up' these contributions along the curve. In physical terms, if \textbf{A} were to represent a velocity field, then this line integral could represent the total distance traveled.
Surface Integral
The surface integral is somewhat analogous to the line integral, but it operates over two dimensions. It measures the total 'flow' of a vector field \(\mathbf{B}\) through a surface \textbf{S}. To determine the magnetic flux through \textbf{S}, you calculate the surface integral of the magnetic field over the surface.

If we return to our equations, the magnetic flux \(\Phi\) is defined as the surface integral of the magnetic field, written as \(\Phi = \int_{S} \mathbf{B} \cdot d\mathbf{S}\). Each infinitesimal area element \(d\mathbf{S}\) contributes to the total flux depending not just on the size of the area, but also on its orientation relative to the magnetic field lines. Hence, the surface integral encompasses both magnitude and direction, providing a complete picture of how the field interacts with the surface.

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Most popular questions from this chapter

Solenoid field, inside and outside *** Consider an infinite solenoid with circular cross section. The current is \(I\), and there are \(n\) turns per unit length. Show that the magnetic field is zero outside and \(B=\mu_{0} n I\) (in the longitudinal direction) everywhere inside. Do this in three steps as follows. (a) Show that the field has only a longitudinal component. Hint: Consider the contributions to the field from rings that are symmetrically located with respect to a given point. (b) Use Ampère's law to show that the field has a uniform value outside and a uniform value inside, and that these two values differ by \(\mu_{0} n I\). (c) Show that \(B \rightarrow 0\) as \(r \rightarrow \infty .\) There are various ways to do this. One is to obtain an upper bound on the field contribution due to a given ring by unwrapping the ring into a straight wire segment, and then finding the field due to this straight segment.

Field in the plane of a ring ** A ring with radius \(R\) carries a current \(I\). Show that the magnetic field due to the ring, at a point in the plane of the ring, a distance \(a\) from the center (either inside or outside the ring), is given by $$ B=2 \cdot \frac{\mu_{0} I}{4 \pi} \int_{0}^{\pi} \frac{(R-a \cos \theta) R d \theta}{\left(a^{2}+R^{2}-2 a R \cos \theta\right)^{3 / 2}} $$ Hint: The easiest way to handle the cross product in the BiotSavart law is to write the Cartesian coordinates of \(d l\) and \(\mathbf{r}\) in terms of an angle \(\theta\) in the ring. This integral can't be evaluated in closed form (except in terms of elliptic functions), but it can always be evaluated numerically if desired. For the special case of \(a=0\) at the center of the ring, the integral is easy to do; verify that it yields the result given in

Equal magnitudes Suppose we have a situation in which the component of the magnetic field parallel to the plane of a sheet has the same magnitude on both sides, but changes direction by \(90^{\circ}\) in going through the sheet. What is going on here? Would there be a force on the sheet? Should our formula for the force on a current sheet apply to cases like this?

Field at the center of an orbit \(*\) An electron is moving at a speed \(0.01 c\) on a circular orbit of radius \(10^{-10} \mathrm{~m}\). What is the strength of the resulting magnetic field at the center of the orbit? (The numbers given are typical, in order of magnitude, for an electron in an atom.)

Motion in \(E\) and B fields \(* * *\) The task of Exercise \(6.29\) is to show that if a charged particle moves in the \(x y\) plane in the presence of a uniform magnetic field in the \(z\) direction, the path will be a circle. What does the path look like if we add on a uniform electric field in the \(y\) direction? Let the particle have mass \(m\) and charge \(q\). And let the magnitudes of the electric and magnetic fields be \(E\) and \(B\). Assume that the velocity is nonrelativistic, so that \(\gamma \approx 1\) (this assumption isn't necessary in Exercise 6.29, because \(v\) is constant there). Be careful, the answer is a bit counterintuitive.

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