Magnetic Flux= BA is A the surrounding area affected by the flux or the area of the coil itself that is affected by the flux
B is measured in Vs/m^2 i.e. flux per area. A can be any arbitary area. btw. The total flux inside a coil is identical in magnitude to the flux outside. Look at the images at http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html The field goes in a circle. Just like the current in a circular circuit has to be the same everywhere, the magnetic flux in a circular magnetic circuit has to be the same.
I mean if a I straight wire with no current is placed in a magnetic field, how can I calculate the Flux affecting the wire using flux=BA
In any calculation with the flux being [tex] \Phi \equiv \int \mathbf{B} \cdot d \mathbf{a} [/tex] it's always best to integrate over the area of the loop of wire. If you have a straight wire, there is no flux to calculate. If there is no current, or if the wire is not moving in the magnetic field, then the magnetic field has no effect on the electrons in the wire because of the Lorentz force law: [tex] \mathbf{F}_{mag} = Q(\mathbf{v} \times \mathbf{B}) [/tex]
If you have a round, single-turn coil with a smll gap in it somewhere, and a time-varying B field is present in-line with the coil's normal vector, then there is also no motion nor current, yet there is an emf developed across the gap. So this argument is questionable. Not that I know the answer. If I have a rectangular path in air but with the wire forming one of the four sides, there is similarly emf developed around that path with the same time-varying B field. The question is, what portion of the emf is along the wire?
What I want to know is If I have a straight wire placed perpendicularly to a magnetic field, how can I calculate the flux affecting the wire " the wire is still, there is no motion"
Oh I was only referring to a constant magnetic field which has no effect. Of course a changing magnetic field induces an electric field. Any current induced in an open circuit by a changing magnetic field would be almost instantly gone as the charge piling up at one end would create an opposing electric field.
The equation [tex]\Phi \equiv \int \mathbf{B} \cdot d \mathbf{a}[/tex] is simply a more specific definition than the one you gave; [itex] \Phi = B \cdot A [/itex] which is really just a special case. The equation [tex]\mathbf{F}_{mag} = Q(\mathbf{v} \times \mathbf{B})[/tex] is saying that the magnetic force on a charge Q is proportional to the strength of the magnetic field and the velocity of the charge. If a wire isn't moving or there is no current, then there is no magnetic force. If you have a loop of wire lying in a plane perpendicular to the magnetic field, then the flux is the magnetic field times the area created by the loop. But only a changing flux will induce an emf in the loop. Like I said in #8, any current in an open circuit wouldn't last long at all as it would almost instantly go away. Now, in general the flux of the magnetic field is BA. If you were to connect both ends of the straight wire within the magnetic field, the area of the loop is what you would use to calculate the flux through the loop. If the wire is not in a loop but open straight, I don't see the point in calculating the flux. I honestly haven't come across this problem, but I assume it's because you aren't going to get any current out of it, at least not for long. None the less, when calculating the changing flux, the area created by the circuit is used to calculate the EMF which in turn can tell us the current. If we have a loop with a small gap, and small enough for current to jump the gap, I would use the area created by the wire and line between the two ends of the wire, treating the air gap as a resistor. If the wire were slightly bent into a U shape, it may or may not be valid to use the area created by the wire and the shortest line between the ends of the wire.
I mention that point, because I couldn't understand how does magnetic flux changes when moving the wire where it cuts the field lines as the wire covers a certain area, it seams non logical to calculate the flux using the area covered by wire !!!
The magnetic flux may not change when moving any wire. By moving the wire, you are setting the charges in motion with respect to the magnetic field, so even if the magnetic field is not changing, the magnetic force is acting on the charges.
If the magnetic field is constant, then as you move the wire, the flux is constant and does not change.
Whenever there is a B, an l and a v. In your case there is no v. So the emf = 0 irrespective of the time behavior of B.
If the wire is moving, then the wire sweeps out an area. Therefore the area is changing, then you do have a changing flux.
Then you use the Blv law. The Blv law applies even in situations where there is moving media and Maxwell's equation del x E = - dB/dt does not apply. In your case the area is formed by the motion of the wire: delta area = lv per second, so dA/dt = B dA/dt = Blv.
I can't understand how changing area changes flux since the geometrical shape of wire is constant so the flux should be constant!!