5/7 Magnetic Fields

We were asked to draw arrows that indicated the direction of the magnetic field at different points on this metal bar.

This is our representation of how the magnetic field looks around the metal pole. We found that we could use our old formulas that had Electric Field in it and replace it with a new variable B, the magnetic field and we could replace charge, q, with the number of poles, p, . Our old formula with electric field is charge / epsilon but since the net amount of poles in the magnetic field is 0, we can say that the integral of BdA = 0.

This is a visualization of how the magnetic field behaves around the metal bar. We see that the magnetic field looks like the patterns on a pumpkin except on the top and bottom where the arrows don't circle back to the metal bar and keep going.

We were asked to make a prediction of how the piece of metal in the middle of the magnet would behave once it was charged. We found that it didn't matter if the charge was negative or positive, only the direction of the current mattered. 

We were given a problem that asks what the force is on each direction of the metal plate if it lies on the xy-plane and a magnetic field is going straight through it on the z-axis. We found that since the magnetic field is perpendicular all force vectors on the metal plate, the net force was 0N.

We derived a formula to find Force and found that force = qV x B. Since we know that current = charge / time and velocity= length / time, we can interchange q and v with I and L depending on the given information. 

This is an example of how a Cathode Ray Tube behaves when their is a strong magnetic attraction/repulsion near the tube. Because the magnetic force from the magnet is strong, the electrons inside the tube are attracted/repelled toward/away from the magnet. Originally, the green dot was at the center of the screen but because of the magnet, it has shifted slightly to the left.

We draw a diagram of how the inside of a Cathode Ray Tube works and found the resultant force vector if there were a magnetic field being directed at the electrons from a certain direction. We used the right-hand rule to determine the direction of the force and found the dots displacement if it started at the center of the screen.

We were given a problem to find the acceleration of a proton in a magnetic field with the given information. 

This is 2 different problems that we were asked to find the force of. In red, the metal plate is parallel to the direction of the magnetic field and we found that the force on the top and bottom were 0N and the force on the left and right canceled each other out resulting in a net force of 0N. In green, we were asked to find the torque of the metal plate if the magnetic field is going through the z-axis and the plate lies in the xy-plane. We found that the force on the left and right were 0N and the force on the top and bottom were going in opposite directions. This caused the plate to spin clockwise. 

We were given a problem that asked what the different forces were at 15 different segments of a half circle. We were given the radius, magnetic field, and the current. We found that at the rightmost and leftmost parts of the half circle, the force was 0N and the force is greatest at the center. This makes sense because the magnetic field is tangential to the force vector at the left and rightmost points and is perpendicular at the center. 

5/5 Oscilloscope

This is the Cathode Ray Tube we examined in class. We were told that inside the tube, there are 4 plates that help guide the electrons to the end of the tube. The material the end of the tube is made out of turns the fired electrons into a color we can visually see, green. The switch Mason has his finger on changes the orientation of the plates inside causing the electrons, or the green dot, to shift positions.

This is with the Cathode Ray Tube on. We see that the electrons have been fired to the center of the Cathode Ray Tube.

We were asked to predict how the Cathode Ray Tube on the Oscilloscope would change based off a change in voltage from a DC supply, we said the output would be shifted upwards and were correct because the y-axis is in units of voltage and adding voltage would just increase the y-intercept. We then derived formulas for a velocity of an electron given time and length of the magnetic field. We found that using kinematics, we could find that the velocity of the electron is simply the length/time

We used speakers to help determine how frequency sounds at different hertz. We found that the older one gets, the smaller the frequency one can distinguish. 

The above pictures are from our lab with the Oscilloscope.

This was the Mystery Box part of the lab that we came in on Friday to complete. We found that the yellow plug was just for grounding so when whenever we had a combination of two plugs consisting of yellow, the voltage and wave shape was the same as the combination without yellow. 

These were are conclusions as to what the different plugs were.

4/30 Charge on Capacitors

We set up a closed circuit to charge a capacitor. We then used the charged capacitor to light a bulb without the batteries in the circuit. We measured how long it took for the bulb to go completely unlit and graphed our results.

The above two graphs are of the potential electric energy of the capacitor vs. time. The first graph is when the capacitor is discharging and the second graph is when the capacitor is charging. 

Using the data we found from the previous lab, we derived a formula that could calculate the charge if given time, capacity, and resistance. The formula we derived in red matches the format of the best fit curve on our graphs. We also found the formula for the time constant.

We found that current has an inversely proportional relationship with time. We used the formulas we derived from the experiment to assist us in solving the problem. We were asked to find the amount of time it took the current to go through the resistor.

We then calculated how long it would take for the circuit to charge a single electron. We replaced V with Q/C so that we could solve the problem and found that it takes 148.7 seconds to charge an electron with this circuit.

This is a demonstration of what happens to capacitors when they are overcharged. We hid behind a blast shield because the explosion is dangerous. Mason also told us that he had experienced a time during the robotics competition when a student had failed to wear protective gear and was injured by a similar explosion. This was in a way to teach us the values of being safe.

4/28 Capacitors

After our pop quiz, we did the problem in class to find the correct solution. First, we had to make an assumption on the direction of the current. We found that since the bottom battery had a higher voltage than the battery in the middle, the middle battery would be overtaken and act basically as a resistor.

The above 3 pictures is from our experiment of the relationship between capacitors and the distance between the two plates. We found that the closer the plates, the higher the capacity. This made our graph an inverse function. The graph is shifted slightly over to the right to compensate for the fact that at 0 distance, the capacity is 0 because it forms a closed loop.

We measured the surface area of 2 pieces of thin foil and separated the 2 pieces of foil by a certain amount of pages. We measured the thickness of the pages and calculated the capacity of the capacitor formed by the foil. We found that we would need 3.55 miles of foil to replicate this exact experiment to yield 1 farad. 

We visualized the previous experiment by drawing it on the left. We calculated the capacity of 2 capacitors when they are arranged in series and when they are arranged in parallel. The bottom right is a separate problem. We were asked to find the equivalent capacity, the voltage of the battery, the total energy, and the charge at point 1 and point 2 of the circuit drawn in purple. 

4/21 Current in a Closed Circuit

Professor Mason asked us what would happen when the switch in the middle of the circuit was closed. The picture on top is the circuit. The picture below are our predictions on how the light bulbs would react when the switch was closed. We predicted that the top bulb would become brighter and the bottom bulb would become dimmer. Since the potential difference between the bulbs was 0, the bulbs stayed the same. 

This setup is how we were able to produce the brightest bulb using the provided material. We also did a setup of when our bulb was dimmest (not in picture). Using the multimeter, we determined that the brightness of the bulbs depends on the current, voltage, and power.

Using the data we collected from the previous lab, we filled out the tables and questions in the lab manual.

We were asked to find the equivalent resistance of the 4 circuits. We found that in series, the equivalent resistance is the sum and in parallel, the inverse of the equivalent resistance is equal to the sum of the inverse of each resistor.

Similar to the previous exercise, we look at an entire circuit to find the equivalent resistance of each circuit. Going from left to right shows our diagrams simplifying and equivalent resistance is on the bottom left.  

We found the resistance of 4 different resistors on the left. The left side of the table is the resistance of the resistors by decoding the colors indicated on the resistors. On the left, we calculated the resistance using the multimeter. We found that for all 4 resistors, the resistance found using the colors was within uncertainty of their actual resistance. Even though the resistors did not match exactly, it was within uncertainty.

vPython Activity

Here is the vPython activity that we were asked to do at home. The only main difference between this and the provided code was that this one required a loop. The condition was that the loop could only run when the loop is less than 2pi. I incremented it at pi/18 so that it would display a total of 36 electric charges. 

4/16 Electric Potential

We derived equations that would calculate the electric potential at different points. We calculated first when the point is a distance x from the center of the ring. We found that we could use the radius and the distance x to calculate the distance from the point to the ring at all parts of the ring. We then calculated the electric potential at a point x away from the top of the ring. This was more difficult because we did not have a constant distance away from the ring at all points on the ring. We had to use trig to come up with a way to sub in for the distance from the point to the ring.

We calculated for the electric potential a different way by using the integral of E * dA. We calculated for the electric potential of when the point is a distance x from the center of the ring and found that we ended up with the same answer as the one we did when he used V = kq/r to derive the answer.

We did the same problem except we used excel to calculate the numerical values of the electric potential at each segment of the ring. Since there were 20 segments, we came up with 20 different values for the electric potential at each segment of the ring. 

We are given a a problem of a point that is above a rod (not above its center) and were asked to find the electric potential of the point on any part of the rod. The distance from the point to the rod was not uniform and ended up with a radical in the denominator that could not be integrated without using integration techniques.

We did an experiment with a multimeter to find the difference in electric potential from a negative charge to another point on the conductive paper. We used the voltage meter to see that the electric potential between the negative and positive was 15V. We then grabbed the two ends of the wire and put them on the conductive paper to see what the electric difference was when the two points were at different distances.

We recorded the electric potential at the different points on the table to the right and answered the questions.