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 лайка
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 лайка
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.
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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 лайк
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 = \hat z\int\frac{\mu_0}{4\pi}\frac{\vec j dV\times(-\vec r)}{r^3}=\int\frac{\mu_0\rho\omega r\sin\theta\cdot(\hat\varphi\times(-r\hat r))\cdot\hat z}{4\pi r^3}r^2dr\sin\theta d\theta d\varphi=
=\frac{\mu_0\rho\omega}{4\pi}\int_0^R\int_0^\pi\int_0^{2\pi}r\sin^3\theta dr d\theta d\varphi=\frac{1}{3}\mu_0\rho\omega R^2
Hence, 2C = \displaystyle\frac{1}{3}\mu_0\rho\omega R^2. So…
The answer is
B(x) = \frac{\mu_0\rho\omega R^2}{3}\left(1-\frac{6x^2}{5R^2}\right).
Note that this answer is also consistent with the x=R case. As you remember from the general solution for \vec A,
\vec A =\left(C_1r -\frac{\mu_0\rho\omega r^3}{10}+\frac{C_2}{r^2}\right)\sin\theta\cdot\hat\varphi.
For the |\vec r|>R case, \rho=0, C_1=0 (because |\vec A| is zero on infinity). What we now have is that outside of the sphere, the system acts like a dipole \vec m =\displaystyle\frac{4\pi\rho\omega R^5}{15}\hat z, even when the distances are not significantly greater than R. Anyway, the vector potential produced by this dipole equals
\vec A =\frac{\mu_0}{4\pi}\frac{\vec m\times\vec r}{r^3}=\frac{\mu_0\rho\omega R^5\sin\theta\cdot\hat\varphi}{15r^2}.
So comparing with the general solution gives \displaystyle C_2=\frac{\mu_0\rho\omega R^5}{15}, so \vec B=\nabla\times\left(\displaystyle\frac{C_2}{r^2}\sin\theta\cdot\hat\varphi\right)\Bigg|_{x=R}=\displaystyle-\frac{\mu_0\rho\omega R^2}{15}\hat z is again constistent with the particular x=R case of the \vec B(x<R) equation.
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thanks man
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where would you rate the difficulty of the problem? and plz provide a sol for 3rd part
I was told that this problem is originally Chinese, and I think the 2nd part requires skills that are out of the IPhO syllabus (or there may be a slick solution, idk).
I’m sure you can do the rest on your own bcz the rest of the problem is simply a charge’s motion in the EM-field in the xy-plane. Though it’s mostly mathematics, Morin’s classical mechanics book would be helpful there.
1 лайк
ok thanks a ton