Recent content by sahil_time

No i guess, the magnet's movement induces closed electric loops in space everywhere, that is what Maxwell's equation suggests ∇XE = -∂B/∂t. The fact that if a magnet moves back and forth , there will be flux change everywhere, and hence an electric field given by ∇XE = -∂B/∂t. Also...

Thank you everyone :)

Okay, i think the question has been a little vague i guess. Let me rephrase. --An electron, if moved back and forth, emits radiation. This is the basis of antenna theory. --In contrast, any neutral metal will not radiate if moved back and forth because it is neutral and hence does not...

But that is due to change in flux linkage, imagine a magnet in absolute free space. If i were to then move it back and forth, will it emit radiation just like an antenna?

Why is it not used for practical purposes? Is it because it cannot produce high energy radiation?

If i move a magnet back and forth in free space, will it radiate? If it does not then why?

I apologize for the ambiguity, what i meant for P(B|R) is that the first ball drawn is RED, and without replacement, the second ball drawn is BLUE. So what is the probability that the second ball is BLUE given the first ball is RED.

I think i found the answer , it is unusual. Where 1-1+1-1+1... = 1/2 http://en.wikipedia.org/wiki/Grandi%27s_series#Heuristics So the original RHS = 1 -2 + 2 - 2 + 2... = 1 -2( 1 - 1 + 1 - 1...) =1 -2( 1/2 )...Using Grandi Series =1 -1 =0 =LHS.

I think i probably got it figured, Thanks to your replies. Here is my non-formal solution. Imagine a Bag that contains 16 Blue balls and 4 Red balls. B-16 R-4 Now what will be P(R|B|R) ? (This is analogous to my original question) Now Suppose we go the formal way, P(R|(B|R)) = P(R,(B|R)) /...

No, it is more or less clear, but it seems puzzling when read for the first time. Maybe i will get used to it.

P(A|B) implies that we are now in the domain of B where we have to calculate the amount of A. It says that B has taken place, and hence we are in its Domain (Like in Venn Diagrams). Now as long as we are concerned with A, it is directly concerned with B. Now since B has happened, it does...

I was thinking, that suppose P(I am alive|Building Fell) = 0.1 . Now it does not matter what caused the building to fall, could be an earthquake or weak structure or bomb fell and so on. What matters is that the building fell, so that is why intuitively P(A|B|C) = P(A|B). Meaning P(I am...

Context : I was thinking how would we expand P((A|B),C), so using P(X,Y) = P(X|Y)P(Y), yields us P((A|B),C) = P((A|B)|C)P(C) . So i was thinking how to interpret such a thing ?

How do we interpret P(A|B|C) ? is P(A|B|C) = P(A|B)? First of all, do such probabilities P(A|B|C) even make sense? If they do, Then, P(A|B|C) means, that given C has occurred and given that B occurs which depends on C, what is the probability that A occurs which depends on B? Will...

Doc Al, i think i totally misinterpreted when you said " But all frames will see the same effect--if one frame sees the needle deflect, all frames will" I thought at first, that you meant that if one frame has a needle that deflects, needles in all frames will deflect. I am sure you did not mean...