Why don t you study electromagnetism in depth in high school?

Updated on educate 2024-07-29
15 answers
  1. Anonymous users2024-02-13

    It's too difficult. There are many such things in college, and the only way to avoid them is to teach him on his own first. Each part must be understood and pay attention to divergent thinking, and only after laying a good foundation can we continue to learn.

    My experience this year. At the beginning of the school year, I couldn't understand the electrical class, I couldn't read the book, and then I didn't care about it for a few months.

    One day I suddenly understood the first chapter of the asynchronous motor, and it took me a while to understand it all.

    There are three possible problems with electromagnetism. One is that the foundation of mechanics is not good, the mechanics of the freshman year is the cornerstone of high school physics and even your continued study of physics in college, high school physics is a structure that runs through a line, and poor mechanics will affect your entire high school physics. For example, the concepts of functional principles and conservation of energy, if you don't understand them well, you will be very confused when you do those problems of electric field electromagnetic induction.

    These concepts of force and energy are universal, and the examination of the difficulty of the second year of high school will only be higher. The second is that mathematics feels bad, and the multi-geometric thinking of electromagnetism, students who are good at mathematics will be relatively comfortable to learn, and this will be cultivated by brushing up on the questions. The third problem is that the foundation of the circuit (steady current) is not good (the aftershock left by junior high school physics), which is part of the strange thing of high school physics, because the circuit does not involve physical mechanisms, strictly speaking, it is not real physics, so you may feel uncomfortable when learning circuits.

    I was very confused about the circuit at that time, and I summarized dozens of typical questions to get a little better. The solution is to summarize more and take more notes, the more systematic the summary, the better, and the typical examples can be completely understood.

    I think the 53A College Entrance Examination Edition (Teacher's Book) is a very good book, read the example questions in the book once, do it once, summarize it once, reminisce about it, and then tell it to the people you like, and you can completely understand the high school physics problems. The synchronous version of "Over Again" is also very good, the types of exercises are very complete and the amount of questions is very large, and it will benefit a lot to make a good summary.

  2. Anonymous users2024-02-12

    No matter what subject he is in high school, he learns general common sense, and the most general rules cannot be studied in depth, and students do not have the energy.

  3. Anonymous users2024-02-11

    Why don't you study electromagnetism in depth in high school, you have to learn a lot of things, so you have to learn a little shallow.

  4. Anonymous users2024-02-10

    The question in the second ** is a multiple-choice question in last year's Shanghai object grade test, which is a reminiscence version of the question, and I don't know what the original question is.

    If the width of AB is not counted, it is a thin rod, then AB will not be subjected to the force of the magnetic field I, and the magnetic field generated by the current itself will not be forceful to itself, and the answer should be D.

    If AB has a certain width, then the magnetic field generated by the rails on both sides will produce a magnetic field I to the right of AB, and the answer should be B, similar to the principle of an electromagnetic gun.

    Magnets are not Lenz's law or anything like that, and the answer to the latter is.

  5. Anonymous users2024-02-09

    In Figure 2, the magnetic induction intensity tends to increase according to Lenz's law, and the area should decrease to the left in order to alleviate this trend.

    Figure 1 is likely to be negligible?

  6. Anonymous users2024-02-08

    1. When studying high school physics and electromagnetism, observe more daily electromagnetic phenomena and cultivate interest;

    2. Use your brain more, think about some phenomena of things, connect with the phenomena of daily life, apply what you have learned, and then connect with reality;

    3. Buy some high school physics practice books, spend more time doing electromagnetism-type questions, summarize experience, and review them frequently;

    4. When taking high school physics and electromagnetism classes, carefully preview and listen to the class, listen to the teacher's explanation with questions, and make more small summaries;

    5. In your spare time, do more experiments on high school physics electromagnetism, so that you can learn high school physics electromagnetism.

  7. Anonymous users2024-02-07

    1. When studying high school physics and electromagnetism, observe more daily electromagnetic phenomena and cultivate interest;

    2. Use your brain more, think about some physical phenomena, connect with daily life phenomena, learn to close your feet and apply them, and then connect them with reality;

    3. Buy some high school physics practice books, spend more time doing electromagnetism-type questions, summarize experience, and review them frequently;

    4. When taking high school physics and electromagnetism classes, carefully preview and listen to the class, listen to the teacher's explanation with questions, and make more small summaries;

    5. In the spare time, do more experiments on high school physics car positive magnetism, and do more hands-on, so that you can learn high school physics electromagnetism.

  8. Anonymous users2024-02-06

    Just think about it according to Lenz's law.

    First of all, for 1, PQ wants to leave on the right, and there will be a magnetic force to oppose the departure of PQ, so 1 pair for 2 can be judged to be right by using the right-hand rule.

    For 3, also because the area enclosed between PQ and RS is originally fixed, in order to resist the change, RS will be electrodynamically moved to the right to keep the area between PRQS constant.

  9. Anonymous users2024-02-05

    m(p):m(q) 3:4,p,q are expressed as the time of motion and the electric field force exerted, and then the distance from the point of incidence to point o is expressed respectively, and two equations appear to make them equal.

    I haven't done the question for a long time, if it's wrong, bear with me!!

  10. Anonymous users2024-02-04

    The particle moves like a flat throw in an electric field, which is caused by x=vt; y = at available:

    a = 2yv² /x² ①

    and a = qe m

    Substituting m yields: m = qex 2yv i.e. m qx

    So m p mq = 3 4

  11. Anonymous users2024-02-03

    The oil droplet is between the two plates and is subjected to its own gravity mg (downward) and the electric field force f given to it, and since the oil droplet is negatively charged, the electric field force is upward. If the electric field force and gravity are equal, the resultant force of the oil droplet is equal to zero because the direction is opposite, so that the oil droplet will move in a uniform straight line in the direction of the initial velocity (Newton's first law).

  12. Anonymous users2024-02-02

    The answer is indeed ACD.

    This charged droplet is only subject to gravity and electric field force in the electric field, and if the resultant force of the two forces is not equal to 0, then it is either up or down, no matter which one cannot ensure that the force is collinear with the direction of motion, and the object only does linear motion when the direction of acceleration (force) and the direction of velocity continue to be collinear, so it seems that the resultant force of gravity and electric field force is only equal to 0 to ensure that the droplet moves in a straight line. The resultant force is 0, so what is done is a uniform linear motion. A is correct, B is false.

    The potential difference u=ed,eq=mg,u=mgd q,c is correctly positively charged on the plate, the electric potential is high, the negative charge is from low potential to high potential, the electric field force does work, the electric potential energy decreases, w=uq=mgd,d is correct.

    Finally, I would like to talk about the question you asked, even if the resultant force is 0, think about Newton's first law, shouldn't an object keep its original state and move in a straight line at a uniform speed, the original velocity has an upward component, and you don't need to force it to run upward. If it is shot horizontally, it needs an upward force to pull it upwards.

    Feel free to ask, hehe

  13. Anonymous users2024-02-01

    Note that because "injected into an electric field" means that it has velocity. Moreover, in the electric field, the net force of the electric field and the force of gravity is in the vertical direction, and if you want the particles to move in a straight line in the electric field, there is a kind of 2 situation. First, the velocity of the particle is collinear with the direction of the resultant force, which is either a uniform acceleration or a uniform deceleration.

    Second, the particle is balanced by force, that is, it moves in a straight line at a uniform velocity.

    In this case, it is obvious that this is not the first case, then it must be the second case, that is, the particle is balanced by force, and the direction of the initial velocity of the particle into the electric field is the direction of the oblique upward direction.

  14. Anonymous users2024-01-31

    It is emitted in a straight line, so the upward electric field force and gravity on it are equal to the force of gravity, so it can be seen that the total force it experiences in the vertical direction is zero. Moving upwards, the direction of the initial velocity may be oblique upwards.

  15. Anonymous users2024-01-30

    aBecause the electric field force and gravity are equal and balanced, the state of motion remains unchanged due to inertia, and it moves at initial velocity.

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