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  1. fight_club_alum

    Inductor Energy

    I calculated all the required in this problem correctly but can't understand why the sum of energy extracted and absorbed is not zero. They are: >0: 0J 0 <=E<=25: 4t^2 J 25 <= E <= 50: 4t^2 - 0.4t + 10 * 10^-3t J E > 50: 0 J Does that mean that an inductor stores energy even after there...
  2. fight_club_alum

    Inductor efficiency

    I think yes but will there be loss of energy between the conversions that take place magnetic field to current or vice versa? Thank you
  3. fight_club_alum

    Inductor efficiency

    I think from the equation I can say that it depends on the inductor's coefficient itself. Am I correct?
  4. fight_club_alum

    Inductor efficiency

    Hey everyone, I am taking a circuit 1 course in college and was wondering if an inductor is actually efficient. What I mean is that does it produce more current than the current already supplied. I know it keeps the circuit going even after the voltage source is cut, but does that mean that, if...
  5. fight_club_alum

    Determining the voltage polarity of the independent current source

    if I am going through the positive terminal of a component, I will write P = +V I if I am going through the negative terminal of a component, I will write P = - V I Isn't that correct?
  6. fight_club_alum

    Determining the voltage polarity of the independent current source

    It would make a difference if I am calculating, for instance, the power developed and dissipated
  7. fight_club_alum

    Determining the voltage polarity of the independent current source

    Thank you so much for replying I did so and got -5 V (with that number how can I determine the positive and negative ends of the current source's voltage?)
  8. fight_club_alum

    Determining the voltage polarity of the independent current source

    I am so sorry if I am posting this in the wrong forum; it is just not a homework problem, and I can't find the right place - it's more of a study help question.
  9. fight_club_alum

    Magnetic force due on current

    Yes, I understand now what is happening Thank you so much for clarifying; after the first couple of questions, all involved a triangular arrangement I thought these questions can't be solved without the cosine law! Thank you so much again for clarifying
  10. fight_club_alum

    Magnetic force due on current

    does it give that answer if you do so? if Yes, can you tell the difference between this question and that question, please: Three long, straight, parallel wires each carry a current of 10 A in the positive x-direction. If the distance between each wire and the other two is 10 cm, what is the...
  11. fight_club_alum

    Magnetic force due on current

    Aren't those wires parallel, and the length is 50 cm?
  12. fight_club_alum

    Magnetic force due on current

    hey, thank you for replying to my question If I put them that way they won't give 1.5 mT and I don't think we can use the cosine law in that case
  13. fight_club_alum

    Magnetic force due on current

    This is how I visualize the problem (of course I am drawing this as if it is in the z-y axis); I don't know what will be the next step. Anyone please help me. Thank you
  14. fight_club_alum

    Particle in a circular path due to magnetic field

    Oh, I see what I've been doing Thank you and sorry for taking too long
  15. fight_club_alum

    Particle in a circular path due to magnetic field

    Thank you but if I did this to convert ev to j sqrt(2 * e * (3000e) / mass) <---- where e is the charge of a proton and 3000e is the ev to joule conversion I get the wrong answer
  16. fight_club_alum

    Particle in a circular path due to magnetic field

    great! thank you so much Sorry if I confused the whole forum
  17. fight_club_alum

    Particle in a circular path due to magnetic field

    When I do this v = sqrt ( 2 * e * 3000/ mass of the proton) <---- where e is the charge of the proton and 3000 is the ev I get the right answer So normally I should use (ev) not v
  18. fight_club_alum

    Particle in a circular path due to magnetic field

    thank you so the right one should be v = sqrt ( (2 * e * (3000e) / (Mass of proton) ) <--- e is just the charge of the proton
  19. fight_club_alum

    Particle in a circular path due to magnetic field

    So, the equation should be v = sqrt( (2 * q * ev ) / mass)
  20. fight_club_alum

    Particle in a circular path due to magnetic field

    I'm sorry but I don't understand is v = sqrt( (2q(potential)/ mass) If yes isn't ev/electron charge = volt Thank you
  21. fight_club_alum

    Particle in a circular path due to magnetic field

    The answer comes out if I convert the ev to joule and work normally using the kinetic energy equation 1/2 m v^2 = E But what is the mistake when I use the other equation?
  22. fight_club_alum

    Particle in a circular path due to magnetic field

    to convert from ev to v. Am I correct?
  23. fight_club_alum

    Particle in a circular path due to magnetic field

    e is the charge of the electron on the constant list in the calculator, which is the same as the proton without the sign
  24. fight_club_alum

    Particle in a circular path due to magnetic field

    v = sqrt( (2 * charge of proton * 3000/e) / (mass of proton)) v = 1.893986024 x 10^`15 r = ( (mass of proton) * (velocity) ) / ((magnetic field) * (charge of proton)) r = 24715769.68 m Anyone please help
  25. fight_club_alum

    Path diameter difference of two singly-charged ions in a magnetic field

    M1(50,000)/q (0.4) - M2(50,000)/q (0.4) = 0.025 Is there a special charge for singly charged ions?
  26. fight_club_alum

    Particle that moves in an electric and magnetic field

    It seems that I've been studying for so long now that I couldn't notice what I first learned in this course Thank you so much for pointing out my obvious mistake
  27. fight_club_alum

    Particle that moves in an electric and magnetic field

    Thank you for replying Can I please know how can I take it into account?
  28. fight_club_alum

    Particle that moves in an electric and magnetic field

    m = 0.005 q = -70 x 10^-6 c v = 30,000 m/s Since there is no movement vertically Fb = Mg So, q . V . B = mg So, (70 x `10^-6) . (30,000) . B = (0.005) . (9.8) So, B = 0.0233333 or ~ 23 MT
  29. fight_club_alum

    Calculate the magnetic force

    8000 0 0 5 -4 3 F = (5 x 10^-6) * up (sorry cant write a materix here) F = (5 x10^-6) {0i - 24000 j - 32000 k) F = -0.12 j - 0.16 k Mag of F = 0.2 (I feel that there is something wrong in the question; I don't know)
  30. fight_club_alum

    Find the charge of this particle moving in a magnetic field

    I think I understand, now. Thank you so much, but why didn't the question say an acceleration of -8k m/s^2 or acceleration of magnitude 8
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