9

You have the definition of the vector potential. $$\mathbf{B}=\nabla \times \mathbf{A}$$ According to Stokes' theorem this is equivalent to $$\iint_S \mathbf{B}\ d\mathbf{S} = \oint_{\partial S} \mathbf{A}\ d\mathbf{r}$$ where $S$ is any surface area and $\partial S$ is its boundary line. Now choose for $S$ a circle around the $z$-axis. Then the integral on ...


8

One way to do this (though I wouldn't blame you for accusing me of cheating) is to "remember" the identity (in Cylindrical Coordinates): $$\nabla \times \left(\frac{\hat{\mathbf{\varphi}}}{\rho}\right) = \hat{\mathbf{z}}\,\, 2 \pi \,\delta^2(\rho).\tag{1}\label{1}$$ This is reminiscent of the divergence identity (in Spherical Polar Coordinates): $$\...


5

First let build up some intuition by looking at the divergence in 1D. Imagine you have a river which has a very uniform flow so you can approximate the flow as 1 dimensionsal. The flow is also static in time. This particular river flows in the positive x direction so $v(x)$ is positive and assume at a particular location we have that $\frac{\partial v}{\...


4

The problem with the divergence of the fields you wrote is that it is ill-defined in the origin. So whatever you find is valid only if $r\neq0$ and we need to manually add the value of the divergence in the origin. How do we do it? We use the fact that the integral of the divergence in the volume is equal to the flux of the vector field on a surface which ...


3

Contrary to the comments, this problem can be solved analytically without the use of differential equations. Since this is a "homework-like" problem, but one requiring an unusual technique, I will describe the method in a bit more detail than I would otherwise. I will still leave the details of the calculations to you. Let $x = 0$ denote the ...


2

Say $(M,g)$ is a $2$-dimensional oriented Riemannian manifold, and let $\star$ be the Hodge star operator. Given a vector field $F$ on $M$, we define \begin{align} \text{curl}(F)&:=\star d(g^{\flat}F) \end{align} i.e we take our vector field $F$, convert it to a covector field $g^{\flat}(F)$, take its exterior derivative (so we now have a $2$-form) and ...


1

Since the field line indicates the direction of the field, The field at each point must be tangent to the line. In an xy system, dy/dx = $E_y/E_x$. Putting the (+q) at the origin of an xy system you can use Coulomb's law the find an expression for the resultant $E_y$ and $E_x$ as a function of (x) and (y). That gives you (dy) as a function of (dx). I was ...


1

First of all, no reason to be nervous about $(\vec{p}_1\cdot\vec{\nabla})$; just like $\vec\nabla$, it is an operator. $(\vec{p}_1\cdot\vec{\nabla})\vec{A}$ is a directional derivative and can be thought of as the rate of change of $\vec{A}$ as you move along $\vec{p}_1$ (with "velocity" $\vec{p}_1$). Using vector calculus identities can simplify ...


1

I don't have the final answer but I hope this helps with the vector multiplication. First, expand everything (including $\nabla$) out into its unit vector components. I would use spherical coordinates so that $|\vec r|^3=r^3$ and $\hat n = <1, 0, 0>$, which should make life much easier, even though spherical $\nabla$ is a little more complicated than ...


1

You're right to be concerned because at first sight, it is indeed not obvious that one can extend the definition of $\xi$ to yield a smooth vector field on $TM$ which has a similar expression in every adapted coordinate chart. However it is precisely because we're only taking a specific combination of the basis vector fields that this induces a global ...


1

“The value of the 𝑥 component of $\vec{V}$ at the centre of the face 1 and 2 will be different from $𝑉_x$ at the centre 𝑃.” “The $x$ component of $\vec{V}$” And “$V_x$” are the same thing. This just says $V_x$ is a function, so it’s value changes because $x$ changed going from center to a face. (See note about their bad terminology at the bottom of this.) ...


1

Given a vector V = x . i + y . j the derivative dV/dx = i (in Cartesian coordinates). So a derivative of a vector w.r.t. a component being a vector should not surprise you, I think. For tangent-vectors, just think of the velocity-vector along a curve C over some surface S.


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