Faraday's Law - Balloon Problem

[rofldude188]

Homework Statement

A scheme for power generation proposes to use conducting balloons that expand and contract in waterfalls (with adequate insulation). Consider a balloon of volume V at a given time, with water flowing into it with a flow rate F (m3/s). The water falls down into the spherical balloon causing it to expand. The balloon is in a region where the magnetic field is uniform and vertical of magnitude B.

Relevant Equations

$$\phi = B \cdot dS$$

a) Calculate the proposed induced emf along the equator of the balloon. (horizontal
equator), at the moment indicated above.

$$V(t) = V + Ft \implies \frac{4 \pi r^3(t)}{3} = V + Ft \implies r(t) = \sqrt[3]{\frac{3V+3Ft}{4 \pi}}$$
$$\phi = B \pi r^2(t) = B\pi (\frac{3V+3Ft}{4 \pi})^{2/3}$$
$$V_{ind} = -\frac{d \phi}{dt} = 2FB \pi (\frac{3V+3Ft}{4 \pi})^{-1/3}$$

b) Indicate the direction of flow of the current around equator if B points down

By Lenz's Law, since area of equator is expanding that means more flux lines are entering downwards so there must be an induced current counterclockwise to oppose this.

c) The scheme then allows for the mouth of the balloon to be closed and water to leak
out through holes in the bottom when the volume of the balloon becomes 4V. The
initial leak rate is 2.5 F. Calculate the induced emf along a horizontal ring on the
balloon a vertical distance z away from the centre of the balloon

For this part, would the method not be the exact same as part a)? i.e. z = r and we find r(t) and proceed from there in the same manner as above?

d) Calculate the induced emf along the largest vertical circle on the balloon for c).

Not sure what to do here. Any help would be appreciated.

 

[TSny]

a) Calculate the proposed induced emf along the equator of the balloon. (horizontal
equator), at the moment indicated above.

$$V(t) = V + Ft \implies \frac{4 \pi r^3(t)}{3} = V + Ft \implies r(t) = \sqrt[3]{\frac{3V+3Ft}{4 \pi}}$$
$$\phi = B \pi r^2(t) = B\pi (\frac{3V+3Ft}{4 \pi})^{2/3}$$
$$V_{ind} = -\frac{d \phi}{dt} = 2FB \pi (\frac{3V+3Ft}{4 \pi})^{-1/3}$$

This looks correct so far. But the question asks for the emf "at the moment indicated above". So, what value of the time ##t## should you use that corresponds to this moment?

b) Indicate the direction of flow of the current around equator if B points down

By Lenz's Law, since area of equator is expanding that means more flux lines are entering downwards so there must be an induced current counterclockwise to oppose this.

OK if you mean counterclockwise as viewed looking down from above the balloon.

c)

For this part, would the method not be the exact same as part a)? i.e. z = r and we find r(t) and proceed from there in the same manner as above?

Why do you say z = r? Also, note that the radius of the ring will be different than the radius of the balloon.

d) Calculate the induced emf along the largest vertical circle on the balloon for c).

Not sure what to do here. Any help would be appreciated.

Think about the flux through a vertical circle.

 

[rofldude188]

This looks correct so far. But the question asks for the emf "at the moment indicated above". So, what value of the time ##t## should you use that corresponds to this moment?

OK if you mean counterclockwise as viewed looking down from above the balloon.

Why do you say z = r? Also, note that the radius of the ring will be different than the radius of the balloon.

Think about the flux through a vertical circle.

Yeah you're right for the first part I need to plug in t = 0.

Sorry I misread I thought z was along the horizontal axis but I now see it's along the vertical axis. So in this case you would find the new radius $$r_z^2 = r^2 + z^2$$ and then find area of that circle with respect to time and go from there in a similar fashion to part a) right?

Largest vertical circle would have dS element perpendicular to B right so there would be no V_ind right?

 

[TSny]

Yeah you're right for the first part I need to plug in t = 0.

Sorry I misread I thought z was along the horizontal axis but I now see it's along the vertical axis. So in this case you would find the new radius $$r_z^2 = r^2 + z^2$$ and then find area of that circle with respect to time and go from there in a similar fashion to part a) right?

Largest vertical circle would have dS element perpendicular to B right so there would be no V_ind right?

All of that sounds good except for your equation for ##r_z##. Looks like a sign error.

 

[archaic]

##\phi=B.dS## is wrong, ##d\phi=B.dS## is the flux element.