We have a magnet and iron flakes are sprinkled on it, we can see that the lines of the iron flakes match the magnetic fields.
We used a compass to determine the direction of the magnetic field, the Earth's "north" pole is south though, so we had to reverse the arrows.
Afterwards, we drew out some field lines associated with the magnet.
We then calculated the net field lines in and out of some surfaces drawn to enclose 0, 1, and 2 poles.
The net field lines going into a surface is dependent on the number of poles enclosed. If there are an even amount of poles enclosed, there is no magnetic flux. If there is an odd amount of lines enclosed, there is a net flux not equal to zero.
Here we magnetize a paperclip by putting it against a strong magnet.
We use our new magnet to mess with a compass and influence it's direction.
\
We then split the new magnet in two, we find that its properties still retain after the split.
The two new sticks each have a north and south pole after it was split from the original.
As stated, our conclusion from the field lines, there is only flux when an odd, more specifically, one, poles are enclosed in the region.
Relating the Gauss' Law, we state that the integral of magnetic field B dot product area vector A should be zero. Since there are no "mono" poles, the net flux is always zero.
An overview of what we have covered.
Since we can move the magnetic field around, we can influence the electrons on the oscilloscope.
However, the force due to the magnetic field is the cross produce of the charges times velocity cross magnetic field. Here we can see that if we orientate the magnetic field, we can control the force that is produced.
More positions with the magnetic force.
When we shoved the magnet into the screen, we noticed there was no change, Since the cross product of the two vectors parallel is zero, there is no force.
Here we solve for the units of a magnetic field. We isolate the field from our force equation and find that units cancel to Newtons over Coulombs and velocity. But then we simplified it to Newtons over Current Distance.
This unit is defined as a Tesla.
Here we have a problem where we fine the acceleration of a electron in a magnetic field. We found the force and then divided by the mass.
Since the force will change based on the force changing the velocity, we found that it creates a circular motion along the field. We can equate this motion to centripetal acceleration and we find that the values match up. Therefore the motion is related.
Given a period of revolution, we solve for the magnetic field as a function of the frequency.
Here, since we can make current flow, and there is a magnetic field present, we can cause a force to act on the wire.
Substituting in for the charge and velocity, we can rewrite our force equation as a function of current and length vector cross magnetic field vector.
Since v = L / t and I = q / t we can equate the times and solve. We're left with a quantity that we can substitute in.
We pass a current though the wire that is looped, therefore creating a force upon the wire.
We're asked the predict what will happen to the wire when the switch is flipped and current is allowed to flow.
Since the cross product allows us to find the direction of the force, we can find the torque acting on the wire. After some thought, we say that the wire will stop after a turn due to the changing force on the magnetic field.
As the angle approaches pi / 2 the torque slowly goes to zero and the loop comes to a halt.
Here we find the forces acting upon a half loop wire and we find the differential force due to the differential length.
We integrate for the entire loop and find that the total force is just equal to the current times radius times magnetic field.
If F = ma and qV / d = F then:
qV / d = ma
Subbing for a with a equality that has distance, the distances cancel and we're left with an equation for velocity based on the charge, mass, and potential.
By breaking up the integral into 15 segments, we calculate our force at each given point. As conceptually. the endpoints have 0 force and the midpoint has maximum force.
No comments:
Post a Comment