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?
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...
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...
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."...
\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...
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...
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.
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!
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...
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...
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...
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...
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...
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...
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...
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...
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 +...
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...
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...
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...