Spring 2015
Professor Mason
May 12 Class
Magnetized Pin
In class, we had to draw the magnetic field of an ordinary pin and magnetized pin, and this picture attached below shows us that ordinary pin has positive and negative charges arranged randomly close to each other; while the magnetized pin has separated positive and negative charges inside the pin.
Based on the professor's lecture, we found out that when there is magnetic field, there will exist force; magnetic force is defined as an equation of [F = IL x B vector] and torque as an equation of [T = IL^2B]; thus, we could also say that [F = qv x B vector].
In class, we also learned that there are two ways to destroy magnetism in an object, and those are by heating up the object until it reaches certain temperature, and by hitting the object with a hammer. In class, the professor did an experiment with a magnetized pin; he showed us how to eliminate the magnetism in the pin by heating it up with a blowtorch shown in the picture attached below. The second way is to hit the object with a hammer; hitting the object with a hammer will cause a massive vibration to that object and will eventually lose all the magnet inside the object.
In class, we had to do a calculation to find the net force of a current loop and to find net torque acting on the loop. We found out that the definition of torque is R x F; we also found out the bigger loop has bigger torque.
Experiment with Magnet and Power Supply
In class, the professor told us that the things that most likely would make a motor to fail to work are brush, coil, and commuter. Then, we continued doing to the experiment, which involved this thing shown in the picture below.
We needed three batteries and wire attached to this thing in order to get it working. However, before attaching all the stuff needed, we also needed to adjust the direction of the magnet. After all things had been set up, we attached the wires to the batteries in order to make the thing in the middle with copper spun; the direction of where the thing would spin depends on the direction of the magnet (North and South).
We need to determine the direction of the spin based on the magnet. First, we did North-facing magnet on the left and South-facing magnet on the right. The result is shown in the picture attached below. On the other hand, we also needed to do the other way around, which South-facing magnet would be on the left and North-facing magnet on the right. Also, nothing happened when the magnet poles on both sides are the same (North facing North and South facing South). The magnetism in this thing cancelled each other out when they are facing the same sides. 
The next experiment we did was to make the circular copper wire spin. The things we needed for this setup was a power supply, alligator clips, magnetic bar, circular copper wire, and two copper wire holders to keep the copper wire from falling. First, in order to get it working, we needed to attach the alligator clips to the power supply, and set it to 4.5 Volts; after the set up was completed, the circular copper wire would spin as shown in the picture attached below. Before the setup was completed, we had to rub one end of the copper wire with sandpaper 360 degree; for the other end of the wire, we just needed to rub it 180 degree with sandpaper; we had to do this in order to connect the wire with the power supply.
Experiment with Magnetic Pole
In this experiment, the professor put a magnetic pole in the middle, and he placed 5 compasses around the magnetic pole to see which direction the compasses are pointing.
As shown in the picture attached below, the compasses are pointing in circular motion. The directions of the compasses depend on the flow of the current; therefore, by reversing the current flow, it reverses the direction of the compasses as well. Finally, we could conclude that B vector is perpendicular to the current, and B vector is also perpendicular to 1/r, which r is the distance from magnet to compass.
Loop Calculation
This time, we were given a setup of a loop, and we needed to find the dB, which dB has a definition in a form of equation of [dB = (Uo/4pi) (IdI x r vector/ r^2)].
In this problem, we also found out a new equation, and that is [Fb/Fe = Eo Uo V^2], which could be simplified as [Fb/Fe = V^2/C^2]. In this calculation, we were given that Uo and Eo are constants, which are [Uo = 1.25x10^-6] and [Eo = 8.85x10^-12]. The complete calculations are shown in the picture attached below.
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