# Rotating ball and lorentz force

Please help in this problem. I am unable to solve the 3 parts. Please provide a difficulty rating for this as well, if possible.

Normally, I would calculate B(x) through calculating the magnetic dipole moment, but the x<r condition makes me doubt about it. Calculating through Biot-Savart law would be kinda messy too.

4 симпатии

yeah… I have no ideas…

Maybe we can use the electromagnetic field transformation? In the rotating frame of reference where the sphere is at rest, \vec B'=0 and \vec E' = \vec E. The fields \vec E and \vec B in lab frame and \vec E' and \vec B' in rotating frame have the following non-relativistic relations:

\vec E=\vec E'-(\vec V\times\vec B')=\vec E',\\ \vec B = \vec B'+\frac{1}{c^2}(\vec V\times\vec E').

So if we take \vec V=\vec\omega\times\vec x=\omega x\hat y, and \displaystyle\vec E'=\frac{\rho x}{3\varepsilon_0}\hat x, then

\vec B = -\frac{\mu_0\rho\omega x^2}{3}\hat z.

I didn’t try to use this method before, so I’ll try to apply this algorithm to some other well-known problems to see whether this answer is true or not.

4 симпатии

It seems like the solution is wrong, as these transformation formulas are for the inertial frames of reference, and the rotating frame is not inertial.

1 симпатия

oh i see

any progress, bro?

Dude, I tried to use Biot-Savart law yesterday, but integration by r via online calculator gave me this:

\dfrac{-3x^2\cos\left({\varphi}\right)\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)\left(\cos\left({\varphi}\right)\sin^2\left({\theta}\right)-\cos\left({\varphi}\right)\right)\ln\left(\left|\sqrt{1-\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\sqrt{-\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2+x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-x^2}+\sqrt{\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1}\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)\right|\right)-3x^2\cos^2\left({\varphi}\right)\sin^5\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)\ln\left(\left|\sqrt{1-\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\sqrt{-\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2+x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-x^2}+\sqrt{\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1}\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)\right|\right)-3x^2\cos\left({\varphi}\right)\sin^4\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^\frac{3}{2}\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^\frac{3}{2}\sqrt{\frac{\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2}{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}+1}-x^2\sin^2\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^\frac{3}{2}\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^\frac{3}{2}\left(\cos\left({\varphi}\right)\sin^2\left({\theta}\right)-\cos\left({\varphi}\right)\right)\sqrt{\frac{\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2}{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}+1}}{\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1}+\dfrac{3x^2\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^2\ln\left(\left|\frac{\sqrt{\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1}\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)}{\sqrt{-\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2+x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-x^2}}+\sqrt{1-\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\right|\right)-3x^2\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^2\ln\left(\left|\frac{\sqrt{\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1}\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)}{\sqrt{-\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2+x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-x^2}}-\sqrt{1-\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\right|\right)}{4\left(\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1\right)}+\dfrac{x^2\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^2}{4\left(\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1\right)\left(\frac{r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)}{\sqrt{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\sqrt{\frac{\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2}{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}+1}}+1\right)}+\dfrac{x^2\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^2}{4\left(\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1\right)\left(\frac{r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)}{\sqrt{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\sqrt{\frac{\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2}{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}+1}}-1\right)}+\dfrac{-x^2\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^2\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)-x^2\cos^3\left({\varphi}\right)\sin^5\left({\theta}\right)\left(\cos\left({\varphi}\right)\sin^2\left({\theta}\right)-\cos\left({\varphi}\right)\right)\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)+3x^2\cos\left({\varphi}\right)\sin^3\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)\left(\cos\left({\varphi}\right)\sin^2\left({\theta}\right)-\cos\left({\varphi}\right)\right)\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)+3x^2\cos^2\left({\varphi}\right)\sin^5\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)}{\left(\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1\right)\sqrt{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}\sqrt{\frac{\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2}{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}+1}}+\dfrac{-3x^2\cos\left({\varphi}\right)\sin^4\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^\frac{3}{2}\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^\frac{3}{2}+x^2\cos^3\left({\varphi}\right)\sin^6\left({\theta}\right)\sqrt{1-\cos\left({\varphi}\right)\sin\left({\theta}\right)}\sqrt{\cos\left({\varphi}\right)\sin\left({\theta}\right)+1}-x^2\sin^2\left({\theta}\right)\left(1-\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^\frac{3}{2}\left(\cos\left({\varphi}\right)\sin\left({\theta}\right)+1\right)^\frac{3}{2}\left(\cos\left({\varphi}\right)\sin^2\left({\theta}\right)-\cos\left({\varphi}\right)\right)+3x^2\cos^2\left({\varphi}\right)\sin^4\left({\theta}\right)\sqrt{1-\cos\left({\varphi}\right)\sin\left({\theta}\right)}\sqrt{\cos\left({\varphi}\right)\sin\left({\theta}\right)+1}\left(\cos\left({\varphi}\right)\sin^2\left({\theta}\right)-\cos\left({\varphi}\right)\right)}{\left(\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)-1\right)\sqrt{\frac{\left(r-x\cos\left({\varphi}\right)\sin\left({\theta}\right)\right)^2}{x^2-x^2\cos^2\left({\varphi}\right)\sin^2\left({\theta}\right)}+1}}

I have almost no idea about calculating \vec B .-.

1 симпатия

Broooooooo…

Does this require some special technique we might not know??

So here’s my attempt to find a solution through vector potential (thanks to @Ersultan for this idea):

We need to use Maxwell’s equation

\nabla\times\vec B = \mu_0\vec j,

where \vec j=\rho\vec v=\rho(\vec\omega\times\vec r)=\mu_0\rho\omega r\sin\theta\hat\varphi. So consider a vector field, the curl of which is a magnetic induction field, \vec B = \nabla\times\vec A. The \vec A is a vector potential, which I will be calculating there.

The convenience of this method is that we can use the spherical symmetry around the z-axis. I mean, if you know the Biot-Savart law for the vector potential

\vec A =\frac{\mu_0}{4\pi}\int\frac{\vec j dV}{r},

you can see that it is collinear to \vec j if the direction of the latter is constant. Then apparently, in a rotating sphere, the \vec A field makes concentric circles around the z-axis and is collinear to \hat\varphi. Knowing that

\nabla\times(\nabla\times\vec A)=\mu_0\rho\omega r\sin\theta\hat\varphi,

let’s find a solution in \vec A = F(r)\sin\theta\cdot\hat\varphi form.

Using

\nabla\times\vec A = \left(\frac{1}{r\sin\theta}\left(\frac{\partial}{\partial\theta}\left(A_\varphi\sin\theta-\frac{\partial A_\theta}{\partial\varphi}\right)\right)\right)\hat r +
+ \left(\frac{1}{r\sin\theta}\frac{\partial A_r}{\partial\varphi}-\frac{1}{r}\frac{\partial(rA_\varphi)}{\partial r}\right)\hat\theta +
+ \left(\frac{1}{r}\frac{\partial(rA_\theta)}{\partial r}-\frac{1}{r}\frac{\partial A_r}{\partial\theta}\right)\hat\varphi,

(this thing gets very simple in our case because A_r=A_\theta=0 and |\vec A| doesn’t depend on \varphi.)

we get

\vec B=\nabla\times\vec A = \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\left(F\sin^2\theta\right)\hat r-\frac{1}{r}\frac{\partial(rF\sin\theta)}{\partial r}\hat\theta =
= \frac{2F\cos\theta}{r}\hat r -\frac{\sin\theta}{r}\frac{\partial}{\partial r}(rF)\cdot\hat\theta.

Now let’s take the curl again, using \displaystyle (B_r, B_\theta, B_\varphi)=\left(\frac{2F\cos\theta}{r},-\frac{\sin\theta}{r}\frac{\partial}{\partial r}(rF),0\right) (their derivatives with respect to \varphi is zero):

\nabla\times(\nabla\times\vec A) = \left(\frac{1}{r}\frac{\partial(rB_\theta)}{\partial r}-\frac{1}{r}\frac{\partial B_r}{\partial\theta}\right)\hat\varphi =\left(\frac{2F\sin\theta}{r^2} -\frac{\sin\theta}{r}\frac{\partial^2}{\partial r^2}(rF)\right)\hat\varphi=\mu_0\rho\omega r\sin\theta\hat\varphi.

As now you see, this thing is proportional to \sin\theta, so we made our assumption about A(r,\theta) right. Next we have to solve a second order differential equation:

F''+\frac{2F'}{r}-\frac{2F}{r^2} + \mu_0\rho\omega r = 0

Using variation of parameters, one can get

F(r)=C_1r -\frac{\mu_0\rho\omega r^3}{10}+\frac{C_2}{r^2}.

So, C_2=0 because infinite F(0) (which corresponds to infinite B) makes no sense. And FINALLY the vector potential takes its form as

\vec A = \left(Cr-\frac{\mu_0\rho\omega r^3}{10}\right)\sin\theta\cdot\hat\varphi,

and the magnetic field is

\vec B = \left(2C-\frac{\mu_0\rho\omega r^2}{5}\right)\cos\theta\cdot\hat r - \left(2C-\frac{2\mu_0\rho\omega r^2}{5}\right)\sin\theta\cdot\hat\theta,

and in our B(x, 0, 0) case (r=x, \theta =\pi/2, so \hat\theta=-\hat z):

\vec B = \left(2C-\frac{2\mu_0\rho\omega x^2}{5}\right)\hat z.

Now, there are different approaches to finding C. We can manually find the \vec B at the origin, for instance. It equals to

B_0 = \int_0^R\int_0^\pi \frac{\mu_0(\rho\cdot rdrd\theta\cdot\omega r\sin\theta)\cdot\sin\theta}{2r}=\frac{\pi\mu_0\rho\omega R^2}{8} = 2C.

So the \vec B(x) equals

\vec B=\mu_0\rho\omega R^2\left(\frac{\pi}{8}-\frac{2x^2}{5R^2}\right).

Note that the answer is probably wrong because it doesn’t suit the x=R case when the magnetic field outside of a sphere is

\vec B =-\frac{\mu_0}{4\pi}\frac{\vec m}{r^3},\quad \vec m =\frac{4\pi\rho\omega R^5}{15},\quad\Rightarrow\quad \vec B(x=R)=-\frac{1}{15}\mu_0\rho\omega R^2\hat z,

that is, the field of a magnetic dipole at the origin. I’m quite sure about the \vec A function, but it still seems like I’m missing something. I’ll update this answer when I find the mistake. (and I hope the approach wasn’t a mistake itself.)

7 симпатий

thanks man