<p>Wow... I just saw four study threads next to each other on top of the page, without PHYSICS!!</p>
<p>So, let's discuss physics.</p>
<p>Each poster answer the question asked by the previous one, then pose his/her own.</p>
<p>Question:</p>
<p>What is the speed of light (the equation, not just the number)</p>
<p>I know n=c/v</p>
<p>So that means that c (speed of light) is </p>
<p>nv=c</p>
<p>Also c=(lambda)f ?</p>
<p>I read somewhere that it is 1/sqrt(epsilon * mu), which are the permmitivity of space and permeability of space, respectively.</p>
<p>I just said the one I said because it's the easiest one</p>
<p>Okay, I've got a question. This is directly from the Acorn Guide by CB. "Calculate the magnitude and direction of the force in terms of q, v, B, and explain why the magnetic force can perform no work." </p>
<p>Would it be F=qvbsin(theta) and for the explaination, would it be because when (theta) equals 0 degress, F = 0? (but then what does it mean when it states "no work," shouldn't it be no force)</p>
<p>Added: or would it mean since W=Fd and magnetic force doesn't move, then there wouldn't be any distance covered, making it equal 0?</p>
<p>Also, under what conditions particles will move with constant velocity through crossed electric and magnetic fields?</p>
<p>In addition: Indicate the direction of magnetic forces on a current-carrying loop of wire in a magnetic field, and determine how the loop will tend to rotate as a consequence of these forces.</p>
<p>lol sorry glucose I was thinking about that formula when i posted the question.</p>
<p>genesis: wow that question is hard! i would guess it has something to do with equipotential lines</p>
<p>Which one? Can you answer the other one? Thanks for responding, I was beginning to think I was the only one on this forum.</p>
<p>We haven't gotten to magnetic fields yet ... :-</p>
<p>I'm not sure about either one ><, but just some thoughts..</p>
<p>
[quote]
"Calculate the magnitude and direction of the force in terms of q, v, B, and explain why the magnetic force can perform no work."
[/quote]
</p>
<p>Maybe if the charge is not moving, then there'd be no force and no work? </p>
<p>
[quote]
Also, under what conditions particles will move with constant velocity through crossed electric and magnetic fields?
[/quote]
</p>
<p>When the electric field cancels out the magnetic field, I think. Like, when F=qE = qvBsin(theta), or E = vBsin(theta)</p>
<p>
</p>
<p>I don't think that's possible. The definition of magnetic force relates to when a charge is moving.</p>
<p>Indicate the direction of magnetic forces on a current-carrying loop of wire in a magnetic field, and determine how the loop will tend to rotate as a consequence of these forces.</p>
<p>If you have a loop of wire in a magnetic field pointing, say, to the left of the screen, and the loop is parallel to the screen, and the current is clockwise. Then at the very left of the loop, the force would be pointing out of the screen, while at the right it would be pointing inward, according to the right hand rule. The top and bottom don't matter because they are parallel to the magnetic field. The overall force would then create torque and rotation.</p>
<p>Isn't that Torque on a Current Loop; Magnetic Moment? (From 20-9 of the Giancoli book)</p>
<p>But for some reason it states that as a Physics C topic...but I think you're right. So the direction of magnetic forces would be in and out of the page, therefore rotating it? Also, can you explain how the right-hand rule works in this situation?</p>
<p>Hey I'm using Giancoli too! :) but I think we are skipping some chapters, and I'm not reading up to date, so I don't even know what magnetic moment is...</p>
<p>Well at the left (unit circle with angle 180 deg), the current would be going up, with magnetic field going left: force outward the screen.</p>
<p>At the right (angle 0), current pointing down, magnetic field pointing left, force pointing in.</p>
<p>At angle 90 and 270 the current is parallel to the magnetic field.</p>
<p>For the right-hand rule what if it was in the loop? Wouldn't they cancel out? But I see what you mean, when it's the force outside the loop.</p>
<p>Edit (posted before me :D, must've been within seconds)</p>
<p>:D</p>
<p>but i still have no idea what to do with the first question though..</p>
<p>what's an "acorn book"?</p>
<p>The acorn book is the guide from CB of the objectives that Physics students should cover. Basically it lists out the things that we should know.</p>
<p>No, regardless of the Force exerted (qvBsintheta), the object will undergo a semicircular motion in a magnetic field, which is why work is always 0</p>
<p>new question,
1. consider 2 charges, +q and -q. the distance between the charges is r. let point P be right in the middle of the 2 charges (r/2m from either side). calculate the potential at P.</p>