- #1

Nirmal Padwal

- 41

- 2

- 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.

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.

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