Recent content by microdosemishief

Then for the emf resulting from the eddy current, do i use flemings left hand rule to find that direction?

Then do I use right hand thumb rule to find Thanks. Now for a thin plate, why is the eddy current circular? Is it perfectly circular?

So in 1. is it anticlockwise cuz current induced between points C and B go from C to B and for 2. it is clockwise

Regarding using the RHR, my thumb is in direction of motion (parallel to page), my index finger pointing along magnetic field (into or out of page), which leaves third finger for induced current pointing down like gravity or up - this doesnt seem right…

Does the direction my second finger points in indicate the literal direction of magnetic field at that point in space, or do I need to further use right hand rule or something for a circular path around the finger

I just read the textbook answer but I want to know why my answer is wrong and not just a different method to the same answer

R

Yes and dE = kdQ/r^2 so the denominator is r^2 where r = Rcos(θ) squared, because dE_x is Rcosθ distance away

I wrote that in my question

My attempt: due to symmetry along x-axis, net E is only along x^hat. dQ = λ dl = λ (R dθ) for each dl, the x component of distance from dl to the origin is Rcos(θ) Hence, E_x = \int_{-θ_0}^{θ_0} k(λ R dθ)/(Rcos(θ)^2 = kλ/R \int_{-θ_0}^{θ_0} sec(θ)^2 dθ = 2kλ/R tan(θ_0) along negative x^hat...

Assuming initial displacement is zero, if x = v₁(t) + ½a(t)², why isnt Δx = x2-x1 = (v₁(t2) + ½a(t2)²) - (v₁(t1) + ½a(t1)²) = v₁(t₂ - t₁) + ½a(t₂² - t₁²) Why is it instead v₁(t₂ - t₁) + ½a(t₂ - t₁)²