Magnet and Electricity

We started the day with understanding the difference between unmagnetized object and magnetized. When an object is unmagnetized, it contains a positive and negative poles that are all over the place. When the object is magnetized, the poles are ligned up.

We know that an unmagnetized object can be magnetized, but can it be the other way around? We then pedicted that it can be by using heat, forceful contact with similar pole, or by hitting it with a hammer. Then, we confirm the first prediction by heating a magnetized clip. When we heat it long enough, the clip becomes unmagnetized and is not able to pick up a metallic object anymore.

We learned about the foce in magnetic field in the past lecture, now we learn about torque. Given a diagram as below in green, we want to first predict the force. When current is going to the right, the force is facing towards us and when the current started to go to the left, the force is going away from us. But when the current is parallel to the magnetic field, as we learned before, the force is zero. We knew from phys 4a that torque is r x F. In this diagram, r is half of the given length. Thus, by substituting the equation into the force equation, we found torque to be 1/2 IL(square) x B. The L(square) is Area, so it can also be replaced by Area in the equation.

Multiplication of number of loops, current, and area is also a myu, therefore, we can also use myu x B for a torque equation.

Given known as below, we want to calculate the torque. In a circle, area is pi*r(square), by replacing the area and plugging in the knowns, we find the torque. But we needed to be careful with angle. The angle given is between the field direction and plane of loops, the angle in this equation is the surgace area perpendicular to plane of loop.

A motor has some problems that can fail along the way. Those include the brush in the commutator and the coil. 

Given another problem as below, we have current, nunber of loops, radius, and magnetic field. By plugging in, we can again find the torque as below. The angle in this problem though is 90* because the area is perpendicular to the loop.

We then learn how a motor works with batteries and magnet. The commutator is placed perpendicular to the magnet with different poles placed on each sides. The north pole connected to the positive side of the battery and the south pole connected to the negative side. When the commutator is let go, it keeps on spinning to the clockwise from north towards the south pole as shown below. When the cable is switched for the north to be connected to negative side and the south connected to the positive side, the commutator spins counter clockwise. When the magnets are switched with the same setup, the commutator spins clockwise as the first setup and when the cable switched again, the commutator spins counter clockwise again. When the magnets are on the same poles facing each other or when the magnets are removed, the commutator does not spin at all. This shows that the magnet is needed for the motor to work.

We learned about relationship between current and velocity drift. We now want to find the relationship between current and velocity h. We learn that electric field is multiplication of velocity drift and magnetic field. We also learn that electric field is velocity h per work. By equating these equations, we find velocity drift to be velocity h divided by work and magnetic field. Using this equation, we can replace velocity drift and find current to be multiplication of density, charge, velocity h, and time per magnetic field as shown below.

A pole with a current is connected to a power source. We then place compasses around the pole. At first all compasses pointing north, which is placed in a clockwise manner. When the power is turned on, all compasses pointing south or counter clockwise as seen in the picture below. This shows that ...

Last, we tape a cable on a board with a setup as below. By connecting each ends of the cable to a power, the current goes through the cable from negative to positive.

We then learn that the ratio of force of magnetic field and electric field is multiplication of epsilon, myu, and velocity(square), but we knew that epsilon is inverse of myu and capacitor(cube). By replacing it, we found the ratio to be velocity(square)/capacitor(cube).

Lastly, we learn about ampere's law. Integration of magnetic field and change of length is equal to myu multiplied by current inclosed.