In D and E, ##\theta## is either ##0## or ##180^{\circ}##; in either case the torque vanishes.
Multiplying by 4 won't solve the problem. The issue is that the magnetic field makes a different angle with each of the sides of the loop. The correct way to account for this is by using the...
Unfortunately, no. :)
In your analysis of A, B, and C, you're only accounting for one side of each rectangle--the one labeled with an "I" in the diagram. In fact, in the formula ##| \mathbf{\tau} | = IAB \sin(\theta)##, the angle ##\theta## is meant to be taken between the current loop's...
Reading through your work in post #4, I'd strongly suggest you use the following convention: Let ##p## (normal typeface, without arrow) denote a relativistic four-momentum, and let ##\vec{p}## denote its corresponding *three-dimensional* momentum. I think you're confusing these two objects in...
As you suggest, ##\langle L^2 \rangle## is just a way to get at ##\langle L_x^2 \rangle##, which we need to know in order to compute ##\sigma_{L_x}##.
I'm not sure what you mean by your first question--##\sigma_{L_x}## can be calculated exactly, so we don't need to worry about "##\langle...
What's the objective here? To find ##r(t)## and ##\theta(t)##? If so, the solution is well-known...Landau / Lifshitz "Mechanics" has a good summary if you've never seen it.
Your solutions to (d) and (e) look correct, except for one small detail--I can't tell if the right-hands side of your last equation is ##k_1 Te^{ik_1 L}## or ##k_2 Te^{ik_2 L}##. Obviously, one is correct and one is not. :)
If ##Lk_2 = n \pi##, then ##e^{iLk_2} = e^{-iLk_2} = \pm 1##. This...
The trick here is that ##\sigma_{L_x}## is something we can actually compute directly in this state, using the definition $$\sigma_{L_x}^2 = \langle L_x^2 \rangle - \langle L_x \rangle^2.$$ Specifically, you should be able to compute ##\langle L^2 \rangle## without too much trouble, which gives...
In general, these sorts of problems are much easier to solve using relativistic invariants than with a straight-up, brute-force application of the quadratic formula. Let ##p_A##, ##p_B##, and ##p_C## denote the four-momenta of the three particles. Then ##p_A = p_B + p_C##. Rearranging this...
As others here have noted, you don't seem to fully understand the meaning of ##r## in Coulomb's formula ##E = KQ / r^2##. The ##r## in this problem is something else. (Hint: Think centripetal acceleration.)
You've got another incorrect formula in part (1). Think simpler--the electric field...
Since you already calculated the force as a function of the rod's velocity in part (2), you can just apply Newton's second law in the form ##m\dot{v} = F##. This gives a simple differential equation for ##v(t)##. Once you have the solution, find ##T## such that ##v(T) = 0##--that's your answer...
Still, I think the simple case of a rectangular object submerged parallel to the surface of a fluid might be what the problem is getting at. In this case, most of the light that strikes the block will be refracted, leading to a sort of "invisibility" effect.
Actually, I think the point is that total internal reflection can't occur in this situation.
Picture the object as a block of acrylic (for example) submerged in a fluid. If the object is visible from above the fluid, then there must be a path for a ray of light to travel via refraction...
I think the real question here is "why is Newton's second law invalid at high velocities?" Indeed, it seems arbitrary to declare that a rule like F = ma is valid at low velocities, but not when v becomes large relative to c. In fact, this seems to violate the principle of relativity itself...