Solving Helmholtz Coils: Diagram, Magnetic Field & Current

AI Thread Summary
The discussion revolves around determining the direction of the magnetic field and current in a diagram of Helmholtz coils. The user questions whether the magnetic field should be perpendicular to the plane and aligned with the axis. There is uncertainty about the strength of the magnetic field in relation to the number of coils. The user seeks clarification on these concepts, particularly regarding the relationship between coil quantity and magnetic field strength. Understanding these principles is crucial for accurately analyzing Helmholtz coils.
MitsuShai
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Hi
I'm suppose to find the direction of the magnetic field and the current in the diagram below and I was just wondering if I did this correctly. (red is magnetic field and blue is the current)

Diagram: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0702.jpg


And I have a few question (that are in bold) to answer in the link below.
http://i324.photobucket.com/albums/k327/ProtoGirlEXE/100_0709.jpg

For the first question, I honesty don't know what to say about the magnetic field. Shouldn't the magnetic field be perpendicular to the plane, so it should be lying on the axis.

For the second question, I honestly thought that the magnetic field should be stronger because there are more coils but I'm not sure about this because of the statement italicized after the question.
 
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This is about magnetic fields.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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