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    Calculate flux from a radiating disk

    Actually, something that doesn't make sense to me...why is it that my first attempt at the problem (see original post) gives you the same answer even though we're treating the flux as though it were crossing some spherical region and not a plane?
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    Calculate flux from a radiating disk

    Well thank you so much for the help!
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    Calculate flux from a radiating disk

    My latest thoughts: F = \int{I cos^2\theta dA/\rho^2} dA = 2\pi \rho d\rho F = 2\pi I \int{cos^2\theta d\rho /\rho} = 2\pi I \int{cos^3\theta d\rho /z} = 2\pi I \int{cos\theta sin\theta d\theta} = -I\pi cos^2\theta Or, when evaluated from 0 to pi/2: F = I\pi And I think this is where you...
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    Calculate flux from a radiating disk

    Ok, great. So then I think up to here I'm good: F = \int{I cos^{2}\theta dA / \rho^2} One thing that just occurred to me from my last post is that dA is the infinitesimal area through which the flux is being calculated, in our case, the plane at z=Z. I was treating it as an infinitesimal area...
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    Calculate flux from a radiating disk

    It seems like we disagree right off the bat with the equation for flux? Isn't this how it's defined in radiative processes? Where d\Omega = dA/r^2. dA is an "infinitesimal amount of surface area that is located a distance r from the source and oriented perpendicular to the position vector r."...
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    Calculate flux from a radiating disk

    \rho^2 = r^2 + z^2 2\rho d\rho = 2rdr since z is constant. I think I understand most of your reasoning, but I'm still not sure how to go about the problem. A disk as viewed from some point P that is off axis will appear as an ellipse -- ok, that much makes sense. So that means the projected...
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    Calculate flux from a radiating disk

    Hm, ok. Let's see... F = \int{I \cos{\theta} d\Omega} d\Omega = dAcos{\theta}/\rho^2 F = I \int{cos{\theta} dA cos{\theta}/\rho^2} cos{\theta} = z/\rho F = I \int{(z/\rho)^2 dA/\rho^2} = I \int{(z/\rho)^2 2\pi r dr/\rho^2} = 2\pi I z^2 \int{\rho d\rho / \rho^4} = -\pi I z^2/\rho^2 Not sure...
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    Calculate flux from a radiating disk

    Ok...so is what I did above in calculating the flux not the same? The cos term arrises from the projection into the line of sight direction. I am incredibly confused.
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    Calculate flux from a radiating disk

    Ah, should just be the projected area, right? A*cos(theta)?
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    Calculate flux from a radiating disk

    Ok, fair enough. That being said, I don't think I understand the drawing. I'm not sure what that length (perpendicular from X1 to QX2 is? Thanks for the help by the way!
  11. F

    Calculate flux from a radiating disk

    Ok, I think you're essentially asking me to do this (page 32), right? I don't think I fully grasped the concept of solid angle. It's the differential of the projected area (projected onto the direction of the observer) divided by the radius of a sphere squared. So, above, I was regurgitating the...
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    Calculate flux from a radiating disk

    In general, it would be A/t^2, right?
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    Calculate flux from a radiating disk

    Added some work to my last post. Bumping this in case anyone is able to help further. Thanks!
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    Calculate flux from a radiating disk

    Ah, you mean expressing the solid angle in cylindrical coordinates and in turn the flux? So d\Omega ={z d\phi d\rho}/\rho^{2}? The problem also says that by assuming Z>>r, you can use the small angle approximation, so in the expression for flux can I let cos\theta = 1, and then F is entirely in...
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    Calculate flux from a radiating disk

    Homework Statement Suppose you have a disk of radius r at x=y=z=0 with its normal pointing up along the z-axis. The disk radiates with specific intensity I(\theta) from its upper surface. Imagine the observer plane is at z=Z, where Z is much greater than r. Let I(\theta) = I = constant...
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    Density of holes on a sprinkler for uniform water distribution

    Hi all, My classmates and I have been at this problem for some time, and it doesn't look like we're getting anywhere. We'd really love any help in the right direction! Homework Statement A lawn sprinkler is made from a spherical cap (max angle  = 45\) with a large number of...
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    Electric Field Strength Inside Concentric Uniformly Charged Spheres

    Homework Statement The graph in the figure shows the electric field strength (not the field lines) as a function of distance from the center for a pair of concentric uniformly charged spheres. Which of the following situations could the graph plausibly represent? (There may be more than one...
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    Self-Inductance of an Inductor

    Omega for the cosine function is 280. Got it, thanks again!
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    Self-Inductance of an Inductor

    Ok, thank you! That makes sense. I was over-thinking this one. So, I should be left with dI(t)/dt = -(.48)(1/280) ε = LdI(t)/dt And solving for L, I come up with about 285, which doesn't seem to be the correct answer. The correct answer is 3.65mH. Apparently I can't even use my...
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    Self-Inductance of an Inductor

    Alright, so the maximum value that sin(x) can have is 1. So t would have to be 280. I guess I don't understand how we move from knowing the maximum value of the emf to knowing that sin(x) needs to be a maximum. I'm having a hard time picturing this graphically. I know that the voltage for an...
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    Self-Inductance of an Inductor

    Homework Statement An inductor has a current I(t) = (0.480 A) cos[(280 s-1)t] flowing through it. If the maximum emf across the inductor is equal to 0.490 V, what is the self-inductance of the inductor? Homework Equations ε = -L*di/dt The Attempt at a Solution I would have that...
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    Maximum Charge on a Capacitor in an LC Ciruit

    Excellent, I got it! Thank you so much. I need to remember to just make whatever substitutions as needed...thanks!
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    Maximum Charge on a Capacitor in an LC Ciruit

    Ah, you're absolutely right. So, you're referring to w = 1/(LC)^1/2, correct? And since I'm given f, I can easily solve for w.
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    Maximum Charge on a Capacitor in an LC Ciruit

    I have considered a conservation of energy approach, but even that hasn't led me far... I know that when the capacitor is fully charged, all the energy is electrical, and in the form Q^2/2C. At the instant where there is a charge and a current flowing, the total energy is givey by 1/2LI^2 +...
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    Maximum Charge on a Capacitor in an LC Ciruit

    Homework Statement A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time t = 0 the capacitor is fully charged. At a given instant later the charge on the capacitor is measured to be 3.0 μC and the current in the circuit...
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    Magnetic Field of a Straight Current Carrying Conductor

    Whoops, guess I forgot to include the image. Sorry about that! Anyone, I've got this one figured out. Thanks for the help!
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    Magnetic Field of a Straight Current Carrying Conductor

    Homework Statement The figure shows an end view of two long, parallel wires perpendicular to the xy-plane, each carrying a current I but in opposite directions. Derive the expression for the magnitude of B at any point on the x-axis in terms of the x-coordinate. Homework Equations...
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    Potential Differences on an Electrical Circuit

    Got it! Thanks so much for the help!
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    Potential Differences on an Electrical Circuit

    Let me try this again. When the switch is open, we have V = emf - IR - (1/2)IR. The 1/2 is the equivalent resistance of B and C, right? From there: emf = IR + 1/2 IR emf = 2/3 (IR) So V = 2/(3emf) Somewhere along the way I'm messing up so that emf doesn't appear in the numerator, but I'm...
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