Finding the flux along the curved part of a surface

In summary: I don't think so. I am not sure what you are trying to say.No, I am suggesting that of the two fluxes through the flat surfaces it is supposed to say that one is in and the other... is not.
  • #1
Pablo
16
2
Homework Statement
The figure shows a closed surface. Along the flat top face, which has a radius of 4.5 cm, a perpendicular field of magnitude 0.18 T is directed outward. Along the flat bottom face, a magnetic flux of 0.60 mWb is directed outward. What are the (a) magnitude and (b) direction (inward or outward) of the magnetic flux through the curved part of the surface?
Relevant Equations
Flux on a surface = 0 (by Maxwell's equations)
I have this problems I am trying to figure out:
Screen Shot 2019-11-05 at 3.24.38 PM.png


So I know that $$\int B\, dA = 0 = \phi_{total}$$
$$\phi_{total} = \phi_{bottom} + \phi_{top} + \phi_{side} = 0$$

$\phi_{side}$ must be equal to the other two fluxes, since they are both outwards:

$$\phi_{side} = \phi_{bottom} + \phi_{top}$$
$$\phi_{side} = 0.6*10^{-3} + \pi*(0.045)^2*0.18=1.745Wb$$

However, I am getting that 1.745Wb is the wrong answer. Does anyone know why?
 
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  • #2
Pablo said:
Homework Statement: The figure shows a closed surface. Along the flat top face, which has a radius of 4.5 cm, a perpendicular field of magnitude 0.18 T is directed outward. Along the flat bottom face, a magnetic flux of 0.60 mWb is directed outward. What are the (a) magnitude and (b) direction (inward or outward) of the magnetic flux through the curved part of the surface?
Homework Equations: Flux on a surface = 0 (by Maxwell's equations)

I have this problems I am trying to figure out:
View attachment 252397

So I know that $$\int B\, dA = 0 = \phi_{total}$$
$$\phi_{total} = \phi_{bottom} + \phi_{top} + \phi_{side} = 0$$

$\phi_{side}$ must be equal to the other two fluxes, since they are both outwards:

$$\phi_{side} = \phi_{bottom} + \phi_{top}$$
$$\phi_{side} = 0.6*10^{-3} + \pi*(0.045)^2*0.18=1.745Wb$$

However, I am getting that 1.745Wb is the wrong answer. Does anyone know why?
Your approach looks OK to me.

What did you get regarding the "inward" vs "outward" part of the problem?

For what it's worth, there is a convention that is sometimes used that states outward going flux is positive and flux directed inward is negative. I'm not sure if that applies here though since the problem statement just says "magnitude." (So I'm assuming that the expected "magnitude" part of the answer is non-negative.)
 
  • #3
collinsmark said:
Your approach looks OK to me.

What did you get regarding the "inward" vs "outward" part of the problem?

For what it's worth, there is a convention that is sometimes used that states outward going flux is positive and flux directed inward is negative. I'm not sure if that applies here though since the problem statement just says "magnitude." (So I'm assuming that the expected "magnitude" part of the answer is non-negative.)
So, I actually got the direction of the magnetic flux through the curved part of the surface correctly (inward). However, this shouldn't matter for the magnitude of the flux as long as I treat the flux of the side a different direction from the flux on the top and bottom. I don't understand why I am getting the magnitude incorrectly.
 
  • #4
Pablo said:
1.745Wb is the wrong answer
Try mWb!
 
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Likes collinsmark
  • #5
Pablo said:
So, I actually got the direction of the magnetic flux through the curved part of the surface correctly (inward). However, this shouldn't matter for the magnitude of the flux as long as I treat the flux of the side a different direction from the flux on the top and bottom. I don't understand why I am getting the magnitude incorrectly.
Again, I agree with you. By that, I at least came up with the same answer that you did.

For what it's worth, the reason that inward flux is often treated as negative is because the surface area vector is conventionally pointed in the outward direction. So when you take the dot product between the magnetic field and the surface area vector, the result will be negative on patches where the magnetic field has a net "inward" component.

[itex] d \phi = \vec B \cdot \vec{dA} [/itex]

Although [itex] d \phi [/itex] is a scalar, it can be positive or negative.

I was just throwing it out there that maybe the homework program is expecting a negative value for the magnitude. I don't know. The problem statement did say "magnitude," so I wouldn't think the negative sign is necessary. But, lacking some other explanation, maybe that's what it is expecting.

[Edit: Oop! Haruspex figured it out.]
 
  • #6
haruspex said:
Try mWb!
That's it. :smile: (or bring back the "[itex] \times 10^{-3} [/itex]" that got lost somewhere in the submitted answer.)
 
  • #7
tried doing mWb (both 1.745 and -1.745), but still wrong:

Screen Shot 2019-11-05 at 4.43.52 PM.png
 
  • #8
Pablo said:
tried doing mWb (both 1.745 and -1.745), but still wrong:

View attachment 252411
You are quoting more digits than in the given data. Try rounding to 2 or 3. Beyond that, I give up.
 
  • #9
haruspex said:
You are quoting more digits than in the given data. Try rounding to 2 or 3. Beyond that, I give up.
that shouldn't be a problem because there is a margin of error of 2% for answers.
 
  • #10
Pablo said:
that shouldn't be a problem because there is a margin of error of 2% for answers.
Then we are down to guessing the error in the question. Maybe it was meant to say that both flat faces have a flux upward.
 
  • #11
I also put the negative value, and it was wrong:
Screen Shot 2019-11-05 at 7.05.10 PM.png
 
  • #12
Pablo said:
I also put the negative value, and it was wrong:
View attachment 252414
No, I am suggesting that of the two fluxes through the flat surfaces it is supposed to say that one is in and the other out.
 

Related to Finding the flux along the curved part of a surface

What is flux and why is it important?

Flux is a measure of the flow of a physical quantity through a given surface. It is important because it helps us to understand how much of a substance or energy is passing through a surface, which can be useful in many scientific and engineering applications.

How is flux calculated?

Flux is calculated by taking the dot product of the surface area and the vector field, which represents the quantity being measured. This is then integrated over the surface to get the total flux.

Why is finding the flux along a curved surface different from a flat surface?

On a curved surface, the surface area and the vector field are constantly changing, making the calculation of flux more complex. On a flat surface, the surface area and vector field are constant, making the calculation simpler.

What is the significance of finding flux along a curved surface?

Finding flux along a curved surface is important for understanding and analyzing fluid flows, electromagnetic fields, and other physical processes that occur on curved surfaces. It allows us to better understand and predict the behavior of these systems.

What are some mathematical methods used to find flux along a curved surface?

Some common mathematical methods used to find flux along a curved surface include using parametric equations, surface integrals, and differential geometry. These methods can be used to calculate flux for a wide range of curved surfaces and vector fields.

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