Theoretical magnetic Monopole

I'm trying to solve question B2 on the 2012 USAPHO semifinal exam (last page). The solutions are here on the last three pages.

I don't understand the following line:

The flux through the loop will then be $\Phi_B = \int \vec B \cdot dA = \frac{1}{2} \mu_0 q_m\int sin\theta d\theta$

If I ignore this line and the next one (which follows from this one), I understand everything until I arrive at this:

$dp=$ bla bla bla $(sin\theta)^4d\theta$. (Bottom of the second last page)

Then they integrate.

How do they even derive that first formula, where does the theta come from? I do understand that its just like electric fields and I know how to find the flux if a charge is on the axis of a loop. Theta is a constant for points on the axis of the loop, but is the monopole not on the loop? Furthermore why is it sine and not cosine, the sign component is flat on the loop and should thus contribute no flux, right? The cosine components contribute to flux. Why do they treat $\theta$ like a variable?

• I solved this problem a long time ago to get on the USAPhO team! Brings back memories. – knzhou Jan 17 '16 at 5:28
• The solution is a bit badly written. What they mean is, $\theta$ is the azimuthal angle, and it varies in that integral from zero to the upper limit $\theta_0 = \sin^{-1}(b/r)$. Unfortunately, they call the integration variable $\theta$ and the limit of integration $\theta_0$ the same thing. – knzhou Jan 17 '16 at 5:29
• If you calculate the flux by hand, you'll see that it should be a $\sin \theta$ within the integral. It's true that the perpendicular field is $\cos \theta$, but there's another factor for the area element. – knzhou Jan 17 '16 at 5:30
• For the last part, they are using the limit $\theta_0$ to parametrize the motion. This is not any different than, say, describing a falling particle's motion by the height. – knzhou Jan 17 '16 at 5:30
• Thanks for the help so far, is it too much to ask for a diagram? I kinda need to see where the theta is please – Faraz Masroor Jan 17 '16 at 5:30