@harambe The two stars must remain on opposite sides of the fixed COM. So they always have the same angular velocity even if their orbits are elliptical.
@harambe Yes I think that is correct. You could guess this option because the only difference between them is the factor in front of $\pi^2$. Almost always $\pi$ appears as $2\pi$ so we should expect to see $4\pi^2$.
In practice it might crash into a building or an aircraft, and it will quickly lose energy because of atmospheric drag, since the velocity must be very high.
@Hema Yes go ahead. I think we've finished with the main part of harambe's question.
My actual doubt is why there is any field at all if Ampere's law cannot be applied
Since for Gauss's law if no charge enclosed there is no flux
My textbook just says " for the case of the circular loop it cannot be applied to extract the simple expression for the field at the centre of the loop
However there exist a large number of situations of high symetry where it be conveniently applied"
What exactly is wrong with this situation? And won't a field exist only if the law is followed?
@Hema Amperes Law relates the current crossing the surface to the line integral of the magnetic field around a closed loop which is the boundary of the surface. Here the current appears to be in the surface not crossing it, and the magnetic field at the dotted line is not the same as the magnetic field at the centre of the loop.
@Hema The field is correctly predicted by the Biot Savart Law. I don't see how you are applying Ampere's Law here to obtain the field at the centre of the current loop.
If you show me a calculation I can tell you what you are doing wrong.
But it seems to me that you have a vague idea without any details of how you will apply Ampere's Law.
@Hema If you draw a circle around the loop at any point, passing through the centre of the loop, then there is a current passing through this circle and a magnetic field along the circumference of the circle. Similar to the magnetic field around a straight current-carrying wire.
The problem here is the lack of symmetry. Unlike the straight wire, for which the magnetic field is the same at every point around the circle, here the magnetic field will be stronger where the circle is at the centre of the loop than where the circle is furthest from the centre of the loop.
Yes, because the magnetic field varies around the circle (which is the Amperian loop) we don't know how to do the line integral. Unlike for the straight wire, for which we know (by symmetry) the magnetice field must be the same at every point around the circle, so the line integral is simply $2\pi r B$.
I would like if someone could clarify this issue for me:
When dealing with a current $I$ running in a loop with radius $R$ and looking for the magnetic field in the middle of the loop.
By using Ampere's law, i know that the current $I$ runs through a loop with the same radius $R$ ,we get that:
...
@Hema As with Gauss' Law for electrostatics, to apply Ampere's Law usefully, you need to find a loop around which the magnetic field is constant, or known for part of the loop. EG you can sometimes choose a rectangular loop for which the magnetic field is constant on the long sides and zero on the short sides. With Gauss' Law you look for a surface at which the electric field is constant.
@harambe No it means something else. Total energy is conserved between launch point and orbit, so KE+GPE at Earth's surface equals KE+GPE at the orbital radius.
In that case they will all have the same velocity. Force times time is equal to change of momentum, so applying the same force for the same time will produce the same change in momentum.
And since their masses are all the same that means the velocities will all be the same.
