Magnetic Fields and Magnetic Forces (0507)

Magnetic Fields and Magnetic Forces

In this photo, the professor put iron filings sprinkled around the magnet and we observe that there are circles from north pole to south pole

The bar was labeled N on the left and S on the right. As we observed, the magnetic fields entered the S side and exited on the N side.

On the table was a permanent magnet. Interestingly, when the magnet was taken apart into 2 separate pieces, the pieces still behaved like a magnet with north and south parts.

We imagined that the big permanent magnet was just a accumulation of many tiny magnets. Therefore, when we broke the big one into two. the pieces behaved as two separate magnets. The tiniest magnet is the atom, assumingly

The total flux of the dipole magnet is equal to poles enclosed/ constant.

We trying to find the the magnetic field, Since the field is constant, the oscilloscope only showed a single dot on the screen. 

Here we moved the magnetic bar closer to the screen in many direction. As we moving, the dot on the screen moved in different direction. We used the right hand rule if it made sense or not. The formula here is F=v cross B

Here we were trying to find the unit of B.

Here we were practicing finding the acceleration from the given information. We first found the magnetic force, since magnetic force caused the proton to move in the circle. The magnetic force was actually centripetal force. We equated qvBsintheta= ma to find a.

Right Hand Rule

Finding the r by equating the magnetic force with centripetal force.

Here we can imagine that the magnetic is in the positive j direction. Here we had to use left hand rule because the current is actually moving electrons. When the current is in negative i direction, another word, moving from right to left, The force will be up, string up. Otherwise, string would move downward.

Finding magnetic field by equating magnetic force with centripetal force. Since w= 2*pi* frequency, since r= v/w. we put that into finding B equation

We use right hand rule for proton, and left hand rule for electron.

The current was going in the counter clock wise direction. The net force would be zero in a closed loop.

The net force in the closed loop is always zero, but not the net torque.

Homework question.

Our illustration.

Summary:

     Today, in class, we learned about magnetism and its properties. We saw how the magnetic fields looks like on the magnet.  Then we learned about the right hand rule when it comes to currents and force. We also saw that current, length and magnetic field determine the force of the magnetic field. When it is given magnetic charge by being placed in a horseshoe magnet and cut into half, the magnetic charge is shown by using a compass.

Oscilloscopes and Mystery Box (0505)

Oscilloscopes and Mystery Box

We began the class introduced to a piece of equipment named Cathode Ray Tube.The Cathode Ray Tube is an electrical component in some vacuum tubes that produces a narrow, collimated electron beam that has a precise kinetic energy. 

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.

The oscilloscope showed a straight line because the voltage source is constant (battery). Here we used the y-division is 0.5; therefore, the voltage of the battery was roughly about 1.3 V.

Here we used the current generator connected to the oscilloscope. The lines jumped up and down continuously because the current changed the direction.

Here we input 96Hz to the scope. The interval peak to peak is 5.2 with 2mS. 1/(5.2*2*10^-3) is roughly about 96Hz.

We recorded our lab work on the white board.

The image above is Blue and Black(DC)

The image above is Red and Black (AC)

The image above is from Yellow & Black (Open) and Black & Yellow (Open)

The image above is from Black & Green (DC) and Black & Blue (DC)

Summary:

     In today’s class, we learned how to use the oscilloscope. We use different electronics to measure and visualize voltage changes. We learned to how to observant and to manipulate the settings on the machine to reveal the true wave patterns on screen. We learned many things about frequency and DC/AC circuits when connecting them to the oscilloscope. We find to use oscilloscopes in order to see voltage changes and use a simple amplifier to look at how analog electronics are used as a conversion to boost a weak signal into a sound we can hear

Capacitors (0430)

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.

Capacitor discharge.

Voltage cross fully charge the capacitor.

In this photo, we draw a sketch of the brightness of the bulb and the time when it is placed across a charged capacitor without the battery present. 

It is a non-line graph. at first, the time is 0, the bulb is most brightness, then as time increases, the bulb gets dimer and dimer.

Here, we tried to obtain the 5uF capacitor by combining five 1uF capacitor in parallel.

 

The red curve represents the charging capacitor, voltage goes up over the course of time. While the blue curve represents the discharging the capacitor, the voltage drops over the course of time. The equation we used for the curve fit is y=A^(1-exp(-Ct))+B for both curves. Interestingly, the C are the same for both graphs, which implying that C is a constant, we found that C is actually 1/Tau=1/RC (R, C are constants, therefore Tau has to be constant)

Here we were trying to find the unit of Tau, to prove that Tau is actually R*C.

After we found the unit of Tau, we set up the equation Q/C= I*R. Since I=dq/dt. We can integrate both side dq/q=dt/Tau. The result is the last equation we wrote on the board.

We said that brightness can be understood as power. Since I=dq/dt=d(CV)/dt)= C*dV/dt. We can see that I graph would be the derivative of voltage graph

Finding Q max here. since Q=V*C= 4.5*e^(-t*C). Q will be max  when t=0, which then equals to 0.45 C

Summary:

     In today’s class, we analyze the idea of capacitors and the charge buildup and discharging of capacitors in a circuit. We did many experiments on capacitors to understand how they work, and how they affect a circuit. We know that what happens when connected in series or parallel, and we did experiments at the different types of RC circuits about developing a relationship between their charging and discharging.  At the end of the class we did some excises to calculate capacitance, time constants, and maximum charge of a circuit 

Capacitors and Capacitance (0428)

Capacitors and Capacitance

We did a quiz about the Kirchhoff's law.

We use Kirchhoff's Current Law and Kirchhoff Voltage Laws to find 3 equations to find the voltage, current and power of each line, and find the total power of them.

We begin to learn Capacitance. it can store charges, the unit is F, the symbol is C.In this photo, Q is charges and V is voltage, and we know the equation Q=VC, and we can find the work done by capacitance is W= 1/2*V^2*C

We plotted our data point using excel. The graph turned out to be 1/x curve.

K here is the dielectric constant. The formula Q=K*epsilon_not *E*A derived from E=Q/epsilon_not*A (Gaussian's surface chapter). The K constant represents different materials will behave differently in magnitude.

Here is some assigned problem. We were asked to find the side of the rectangular plate. Information needed was given.

We were given two capacitors. We measured capacitance of each capacitors individually. Then we obtained the results of capacitors in series and parallel circuit. Interestingly, to find the equivalent capacitors, we would have to do the opposite process to find the equivalent resistance

Our work to find equivalent capacitor. First,we noticed that top two capacitors were in series, so we combined them. Then we combined the two bottom ones. Now, we had a total of 3 parallel capacitors. we just needed to add them up.

The 3 circuits are parallel to each other. We first find the combined the two capacitors of the top ones, which were in series. Then we combined that capacitor to the bottom one, which were parallel. We redrew our circuit which just a single battery and a single equivalent capacitor. We then find the Q1= C1*V1. Q1=Q2. Therefore, C1V1= C2*V2. We now have V2 and V1. V3= V1+V2.=Vb

Electric Circuit, Kirchhoff's Law (0421)

Electric Circuit, Kirchhoff's Law

The image show above is electrical circuit. Also porf.Mason asked the brightness of the light bulbs when the switch was closed. The light bulbs theoretically had the same brightness. However, since the lower light bulb was defective, the brightness did not come out as we expect

That is our prediction. Because the voltage difference between two side of bulb is zero, so the bulb doesn't light.

Theoretically, the two bulb have same brightness, because the voltage difference cross the bulb is same.

That's our answer.

That's our set up.

We were filling the table on the lab manual. Our jobs were to find the potential difference for each set up

We using the color band code compare measurement, the measurement almost in the range.

Here we were trying to simplify the combination of resistors by making them as like one equivalent resistor. 

For parallel: R_eq= (1/R1+1/R2+...1/Rn)^-1

For series: R_eq= R1+R2+..+Rn

Kirchhoff's Law applying.

Today we went back to study electric circuit. This time we emphasized on theoretical side rather than just knowing how to set the wire up as we did in the last lab. We learned about the Kirchhoff's rule in circuit along with the process of how to equivalent resistors (we believe that we can apply the same process to find the intensity also).