<p>Hehe I'm so gullible. :D</p>
<p>hopefully, they you will grade my exam :D</p>
<p>I think I'm the most unprepared person for this test on the planet, it's okay though, I would take it in college again no matter what I get</p>
<p>BMod, where did you here that?</p>
<p>My teacher said that motional emf might be on this year's exam. There is not much to motional emf, is there?</p>
<p>E = -(Change in flux)/(Change in time) = -BLv (in certain cases)</p>
<p>And when a metal plate carrying a charge enter a magnetic field, its velocity decreases, because the change in magnetic flux causes a force, or a derease in current. And on its way out of the magnetic field, the metal plate slows down even more, because the change in flux has a different sign and the currect is in the other direction, so the force is in the same direction as is what when the plate entered.</p>
<p>Can anyone explain motional emf and torque?</p>
<p>can someone explain that??</p>
<p>torque = iNBAsin(theta) on a loop (or loops) of wire</p>
<p>i: current
N: # of loops
B: magnetic field
A: area of loop</p>
<p>I believe Lenz's Law is also involved in the situation you proposed</p>
<p>Woot! Magnetic flux stuff. I know it. Yes! =). Please have only little rotation...</p>
<p>"And when a metal plate carrying a charge enter a magnetic field, its velocity decreases, because the change in magnetic flux causes a force, or a derease in current. And on its way out of the magnetic field, the metal plate slows down even more, because the change in flux has a different sign and the currect is in the other direction, so the force is in the same direction as is what when the plate entered."</p>
<p>explain this cuz i thought that magnetic field lines are always perpendicular to velocity for charges. why would there be a current, there'd only be a separation of charge on the plate wouldnt there?</p>
<p>if you can remember faraday's law, lenz's law is just basically that its in the opposite direction of the change.</p>
<p>how much do we need to know about L-C circuits? is it just the limits and tau or do we need to be able to go in depth giving the current, q on each and kinetic and potential energies at specific times?</p>
<p>"why would there be a current, there'd only be a separation of charge on the plate wouldnt there?"</p>
<p>the current would be the charges moving to form that given seperation equalizing the force from the magnetic field and the electric field</p>
<p>I was wondering, in the Physics C exam, can you leave your answers non-simplified and still get them counted right? I know it was this way for the Calculus exam, but a teacher told me he wasn't sure about the Physics exam. This would be free response of course. </p>
<p>Thanks to those who reply.</p>
<p>Interesting question, for radicals I doubt you have to simplify but for everything else, I'm sure it's prefered. I'd also am curious about this.</p>
<p>Good luck to everyone .. on this last day of cramming :D</p>
<p>hmmm do we need to know maxwell's equations? from what i'm hearing i'm feeling much better about this test :D</p>
<p>"It is not necessary to simplify all numerical expressions or to carry out all numerical calculations. You will generally receive most, if not full, credit for answers that contain expressions like sin 40° or ln 2, or that contain symbols for irrational numbers. " - from collegeboard.com</p>
<p>I don't <em>think</em> we need to know maxwell's equations, because I think my teacher said we're going to learn that after the test, but then again, what do I know.</p>
<p>how do you differentiate paramatric equations? there has be one on almost every multiple choice</p>
<p>it gives you y and x and asks to find velocity or acceleration at time t</p>
<p>velocity is dy/dx, so if you do (dy/dt)/(dx/dt), the dt cancels out and you get dy/dx.</p>
<p>acceleration is a bit harder. it's d2y/dx2, so it is d/dt of dy/dx over dx/dt. i.e. derive dy/dx, velocity, normally but divide that by dx/dt at the end to get a.</p>
<p>that might take a while to derive and plug in.
i guess ill skip that question and come back to it later
thank you for finally explaining it to me</p>
<p>umm... i velocity is just <dx dt,="" dy="" dt="">. dy/dx is the slope of the path of the particle, not its velocity. acceleration is the 2nd derivative or <d2x dt2,="" d2y,dt2="">.</d2x></dx></p>