Is Your Induction Solution for Rotating Spheres Valid?

[Nirmal Padwal]

Homework Statement

A spherical shell of radius $a$ rotates about the $z$-axis with angular velocity $\omega$. It is in a uniform induction which is in $xz$ plane and at an angle $\alpha$ with the axis of rotation. Find the induced electric field at each point on the sphere.

Relevant Equations

$1. d\Phi = \vec{B}.d\vec{a}$
$2. E_{ind} = \frac{d\Phi}{dt}$

Let $(r,\phi, \theta)$ be the radial, polar and azimuthal coordinates respectively.

As $\vec{B}$ is confined to $xz$ plane such that $\theta = \alpha$ I assumed $\vec{B}$ on the surface of shell to be $\vec{B} = a\sin(\alpha) \hat x + \cos(\alpha) \hat z \tag{1}$

Surface area element is $\mathrm d\vec{a} = a^2 \sin(\theta)\,\mathrm d\theta\, \mathrm d\phi \hat r$. On converting to cartesian coordinates, $\mathrm d\vec{a} = a^2 \sin(\theta),\mathrm d\theta\,\mathrm d\phi\ (\sin(\theta)\cos(\phi)\hat x)$
or $$\mathrm d\vec{a}= a^2 \sin^2(\theta)\cos(\phi)\,\mathrm d\theta\,\mathrm d\phi \hat x \tag{2}$$

Now what we traditionally do is take the dot product of $(1)$ and $(2)$ and then integrate it over the given surface (in this case the surface of sphere). But what I did was as follows:

As the sphere is rotating about $z$ axis, $\phi = \omega t + \phi_0$ (I assumed $\phi_0$ can be taken as zero as its value should not affect the final answer drastically)
Assume a particular $\theta$, for this $\theta$ we substitute $\phi = \omega t$ in $(2)$ and get $$d\vec{a} = a^2 \sin^2(\theta)\cos(\omega t)\,\mathrm d\theta\,\mathrm d(\omega t) \hat x$$ or $$\mathrm d\vec{a} = \omega a^2 \sin^2(\theta) \cos(\omega t)\,\mathrm d\theta\,\mathrm dt \hat x$$

Now, we know that, flux through $\mathrm d\vec{a}$ is $\mathrm d\Phi = \vec{B}\cdot\mathrm d\vec{a}$ i.e
$$\mathrm d\Phi = (a\sin(\alpha))(\omega a^2 \sin^2 (\theta)\cos(\omega t)\,\mathrm d\theta\,\mathrm dt)$$ or $$\mathrm d\Phi = \omega a^3 \sin(\alpha) \sin^2 (\theta)\cos(\omega t)\,\mathrm dt$$ or

$$\frac{\mathrm d\Phi}{\mathrm dt}= \omega a^3 \sin(\alpha) \sin^2(\theta) \cos(\omega t)$$

But this for a particular $\theta$, we now integrate this for $\theta : 0 \rightarrow \pi$ to obtain $$\frac{\mathrm d\Phi}{\mathrm dt} = \frac{\pi}{2} \omega a^3 \sin(\alpha) \cos(\omega t) \tag{3}$$

Now induced emf will just be, by Faraday's law of induction, the negative of $(3)$

As I did not use the traditional method and this was an even-numbered problem (so I don't have the final answer either), I was not sure if this is solution is valid. Please let me know if it is.

 

[Antarres]

As $\vec{B}$ is confined to $xz$ plane such that $\theta = \alpha$ I assumed $\vec{B}$ on the surface of shell to be $\vec{B} = a\sin(\alpha) \hat x + \cos(\alpha) \hat z \tag{1}$

If you've been given the intensity of this magnetic field(induction), shouldn't it read $\vec{B} = B\sin(\alpha) \hat x + B\cos(\alpha) \hat z$? That makes more sense to me, than putting $B$ to be a projection of radius, or whatever. $\vec{B}$ is uniform, so it doesn't matter if it's on the surface of the sphere or not, it's passing through it, and is everywhere constant, that's what the exercise says.

Surface area element is $\mathrm d\vec{a} = a^2 \sin(\theta)\,\mathrm d\theta\, \mathrm d\phi \hat r$. On converting to cartesian coordinates, $\mathrm d\vec{a} = a^2 \sin(\theta),\mathrm d\theta\,\mathrm d\phi\ (\sin(\theta)\cos(\phi)\hat x)$
or $$\mathrm d\vec{a}= a^2 \sin^2(\theta)\cos(\phi)\,\mathrm d\theta\,\mathrm d\phi \hat x \tag{2}$$

You have here converted $\hat{r}$ into something parallel to $\hat{x}$. Those sines and cosines you added only mean the intensity is modified, but essentially, you said that radial unit vector is parallel to the unit vector in $x$ direction, which is obviously incorrect. The transformation law would be:
$$\hat{r} = \sin{\theta}\cos{\phi}\hat{x} + \sin{\theta}\sin{\phi}\hat{y} + \cos{\theta}\hat{z}$$

The rest of it is wrong because of the above two reasons, but you can proceed in the same manner as you did, the method is fine. You just have to be careful with the transformations.
Also keep in mind that $E_{\text{ind}}$ is a potential induced by this change of flux, not the electric field.

 

[Nirmal Padwal]

Thank you for pointing out those mistakes.

So if I assume $\vec {B}$ to have a magnitude $B$ then
$$\vec{B} = Ba sin (\alpha) \hat x + B cos (\alpha) \hat z $$

Also, as you pointed out, the correct surface area element is
$$d\vec {a} = a^2 sin^2 (\theta)cos(\phi) d\theta d\phi \hat x + a^2 sin (\theta) cos(\theta) \hat z $$

Then $$\mathrm d\Phi = \omega a^3 \sin(\alpha) \sin^2 (\theta)\cos(\omega t)\,\mathrm dt + Ba^2 cos(\alpha) sin (\theta) cos (\theta) $$

Now if I integrate this from $\theta : 0 \rightarrow \pi$ the second term evaluates to zero.

So I essentially get the same result for $\frac {d\Phi}{dt}$ as above, just multiplied by magnitude of induction $B$. Is this correct?

 

[Antarres]

So if I assume $\vec {B}$ to have a magnitude $B$ then
$$\vec{B} = Ba sin (\alpha) \hat x + B cos (\alpha) \hat z $$

No need for $a$ in there, you're projecting $\vec{B}$, so it should just be the same formula but without $a$ in it. Sines and cosines determine the value of projections, radius has nothing to do with that.

Also, as you pointed out, the correct surface area element is
$$d\vec {a} = a^2 sin^2 (\theta)cos(\phi) d\theta d\phi \hat x + a^2 sin (\theta) cos(\theta) \hat z $$

Then $$\mathrm d\Phi = \omega a^3 \sin(\alpha) \sin^2 (\theta)\cos(\omega t)\,\mathrm dt + Ba^2 cos(\alpha) sin (\theta) cos (\theta) $$

Now if I integrate this from $\theta : 0 \rightarrow \pi$ the second term evaluates to zero.

So I essentially get the same result for $\frac {d\Phi}{dt}$ as above, just multiplied by magnitude of induction $B$. Is this correct?

You misplaced the differentials but that may be a typing mistake. Either way, the solution for $\tfrac{d\Phi}{dt}$ is similar, you just have $Ba^2$ in the first term instead of $a^3$, according to the correct formula for projection I mentioned earlier.