# Tag Info

47

The vector $\hat \varphi$ is not defined at the origin, because the coordinate transformation $$(x,y) \mapsto (r,\varphi) = \left(\sqrt{x^2 + y^2}, \arctan(y/x)\right)$$ is singular there. Hence your field $\mathbf B$ is singular at the origin. The theorem that $$\nabla \times \mathbf B = 0 \Rightarrow \oint_C \mathbf B \cdot d\mathbf r = 0$$ requires that ...

19

There already are very good answers so I would just like to give some physical intuition why this vector field is curl-free even though it has non zero circulation. We can make an analogy of the curl with an infinitesimally small paddle wheel in a fluid flow. We think of the vector field as a flow of the fluid and the paddle wheel plays the role of the curl....

13

That formula is valid outside the wire, where $J=0$. Maxwell 's equation says that $\nabla \times B = 0$ there. However there is no scalar field whose gradient is $B$ around the wire. This is the typical case where you have an irrotational field which does not admit a (global) potential. Otherwise the integral of $B$ around the wire would be zero and this ...

9

Ampere's law says that $$\nabla \times \boldsymbol{B} = \mu_0 \boldsymbol{J} + \epsilon_0\mu_0 \frac{\partial}{\partial t} \boldsymbol{E}$$ so "far away from sources" means that the current density $\boldsymbol{J}$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be ...

8

OP's magnetic field $$\vec{B}~=~\frac{k }{\rho}\hat{\boldsymbol \varphi}, \qquad \rho~\neq~ 0,\tag{1}$$ in cylindrical coordinates $(\rho,\varphi,z)$ obeys (in a distributional sense) Ampere's circuital law (ACL): $$\mu_0 \vec{J}~\stackrel{(ACL)}{=}~ \vec{\nabla}\times \vec{B}~\stackrel{(1)}{=}~2\pi k ~\delta^2(x,y)\hat{\bf z},\tag{2}$$ with the current ...

8

This diagram might give you the insight you are looking for: I visualize two segments of current (into the plane of the image) on opposite ends of an arbitrary (off axis) point, where both cover the same solid angle. The total amount of current increases with distance (for the same incremental angle, you "see" more of the surface); the actual contribution ...

8

When $\frac{\partial E}{\partial t}=0$, then the 1st equation is valid. That is, it is for magnetostatics, where currents (and fields) are not time-varying.

7

I think the better way to derive this is to first observe the Biot-Savart law, $$\mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\mathbf J(\mathbf r')\times\frac{\hat{r}}{r^2}\,\mathrm dV'\tag{1}$$ Since $$\frac{\hat r}{r^2}=-\nabla_r\left(\frac1r\right)$$ (your text may derive this, if not you can prove it by starting with the RHS), we can write (1) as $$\... 6 As far as I can remember, the formula you obtain is right. You can make this "problematic" integral disappear by using the following identity, that we will call "curl theorem" :$$\int\vec{\nabla}\times\vec{w}dV = -\int\vec{w}\times d\vec{S}$$To show this is true, we are going to use the divergence or Green-Ostrogradski theorem, namely$$\int\vec{\nabla}...

6

The energy of the magnetic field is the work required to establish a general steady-state distribution of currents and fields. This work is, in infinitesimal form, $$\label{0}\tag{0} \delta W = \frac 1 2 \int (\delta \mathbf A \cdot \mathbf J) \ d^3 x$$ where $\mathbf J$ is the current density. If we are interested in work done on the free (...

6

Like all questions about definitions, the answer is not fundamentally "interesting" because, well, it all just boils down to your choice of definitions. But I would define "magnetostatics" to be the regime in which neither the magnetic field nor the electric field depends on time. The motivation for this definition is that the behavior of classical E&M ...

5

I also dislike when authors claim things to be obvious. If it's so simple, then why not just write it out. Anyhow, regarding this specific case. If you go to the definition of the curl you will see that this is a collection of partial derivatives with respect to position. So to claim that the curl is zero is to claim that the velocity is independent of the ...

5

The identity is correct for all of the infinitely many surfaces (isn't math amazing?). In general you either have a surface you care about in the first place or are in a position to choose the easiest surface.

5

Short vision: The most fundamental is the fact that we get same physics in all inertial frames. In this light, the most fundamental quantity in classical electrodynamics is Charge,$^1$ measured and defined as: $Q=\epsilon_0\oint_S\vec {E}\cdot d\vec {a}$. Magnetic force is the consequence of charge invariance under Lorentz transformation and principles of ...

5

The Biot Savart law is equivalent to Ampere's law without the Maxwell term under the assumption that the charge density has no time dependence. So if we have the usual situation where there are currents producing a magnetic field but no net charge density then the two formulas are actually equivalent. (In the case where there is changing charge density but ...

5

On pages 187 and 188 Jackson explains the reason for this singular term. If you take a dipole whose magnetization is distributed uniformly in a sphere of radius $R$ then one can show that $\int_{r<R} \textbf{B}\: d^3{x} =\frac{2\mu_0}{3}\textbf{m}$ where $\textbf{m}$ is the total dipole moment. As one shrinks the sphere's radius $R\to 0$ the sphere ...

5

Partial answer: if there are no currents, all such magnetic fields must be constant. In the absence of currents, we have $$\nabla \cdot \mathbf{B} = 0, \quad \nabla \times \mathbf{B} = 0.$$ The curl-free condition is equivalent to $\partial_i B_j = \partial_j B_i$, as is clear by writing it in terms of differential forms. As a result, the Laplacian of any ...

5

Why do you claim that $$\int_V \nabla \cdot \frac{{\bf J}({\bf x}')}{|{\bf x}-{\bf x}'|} {\rm d}V' = -\int_V \nabla' \cdot \frac{{\bf J}({\bf x}')}{|{\bf x}-{\bf x}'|} {\rm d}V'$$? This doesn't seem to be true. You should have \begin{align} \int_V \nabla \cdot \frac{{\bf J}({\bf x}')}{|{\bf x}-{\bf x}'|} {\rm d}V' &= \int_V \Big(\nabla \frac{1}{|{\bf x}-...

4

There is no "physical aspect of this fact". The physical variables are the electric and the magnetic field, not the potentials. Introducing the potential is aesthetically and technically pleasing, but it is not necessary. A gauge symmetry is not a physical symmetry. The reason you can have a non-unique potential is that every divergence-free field such as ...

4

In Jackson's textbook, paragraph $\S 5.3$ starts with equation (5.14) given here as (001) \begin{equation} \mathbf{B}\left(\mathbf{x}\right)=\dfrac{\mu_{o}}{4\pi}\int\:\mathbf{J}\left(\mathbf{x}'\right)\boldsymbol{\times}\dfrac{\mathbf{x}-\mathbf{x}'}{\left|\mathbf{x}-\mathbf{x}'\right|^3}d^{3}x' \tag{001} \end{equation} Then using the relation just above (...

4

The condition that the $p_i$ be conserved is equivalent to $\{H,p_i\} = 0$ for the Hamiltonian $H = (p-A)^2$, where I've dropped all constants for convenience. A straightforward computation yields $$\{H,p_i\} = -2\sum_j \frac{\partial A_k}{\partial x^i}(p_k - A_k)$$ and $p_k =A_k(x)$ is impossible since this is an off-shell equation where $p$ and $x$ are ...

4

When you ask why does Ampère's law often apply to an infinitely long wire and the Biot Savart Law apply to a short wire? you're completely mistaken: both Ampère's law and the Biot-Savart law always hold. More specifically, if you have a current $I$ running over a curve $\mathcal C_0$, then: The Biot-Savart law specifies the magnetic field $\mathbf B(\... 4 A first observation is that this is not particular to magnetism. The exact same thing happens if you try to find Coulomb's law for the electric field; you get a term like $$\int \nabla' \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}\ d^3 \mathbf{x}'$$ which should be zero. Well, there are no fancy vector calculus identities involved, just plain old ... 4 No physics, just a purely mathematical argument. Stokes's theorem:$\oint_{C} {\vec{v}}\cdot d{\vec l} = \int\int_{S} curl\vec{v} \cdot \vec{dS}$, where$\vec v$is any differentiable vector field and$C$any simple (piecewise) smooth loop that bounds a simple smooth (etc.) surface$S$. So if Maxwell's equation is written as$\oint_{C} {\vec{B}}\cdot d{\vec ...

4

Just for picturing it...meaning this is not exact: If you are not in the middle, where it is clear, but close to one side you may see it like this: the length over which you see a wire approximately straight is sort of given by the angle. Hence, the amount of wire with approximately opposite direction on the other side increases with the distance to the ...

4

The field inside the sphere is $$\label{1}\tag{1} \mathbf B_{in} = \frac 2 3 \mu_0 \mathbf M$$ where $\mathbf M$ is the magnetization vector. The magnetic dipole moment is $$\mathbf m = V \mathbf M$$ where $V = 4 \pi R^3 / 3$ is the volume of the sphere. I use the notation $\mathbf m$ for the dipole moment because $\mu$ is usually used for the magnetic ...

4

The position vector $\vec r$ is the position of $A$ where you want to find the magnetic field relative to the position of the current element at position $B$.

4

As you say. the Biot-Savart law holds whenever $\dot{\bf J} \equiv {\bf 0}$, even if ${\bf \nabla} \cdot {\bf J} \neq {\bf 0}$ and so $\rho$ changes linearly over time. The assumption $\dot{\rho} \equiv 0$ is not necessary. Moreover, since ${\bf J}({\bf x}, t) \equiv {\bf J}({\bf x}, t_r)$ in this case, we trivially have that that the instantaneous version ...

3

In short, your teacher is right. Permanent magnets (ferromagnetic materials) are divided in regions in which there's a net dipole moment, such that when all of them are added up there's a magnetic field. The dipole moment is not due only to the motion of charges, it is due to the fact that electrons have an intrinsic dipole moment, known as spin, and an ...

3

The magnetic field of a permanent ferromagnet is generated by the individual magnetic dipole moments of unpaired electrons within it. In a ferromagnet the unpaired electrons line up their spins in the same direction so their individual magnetic dipoles sum to give a macroscopic magnetic field. The magnetic field of an electron is a purely quantum effect and ...

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