Recent content by PolarBee

Oh I see~ I was taking into consideration of of the whole rotation of my hand, so the back of my hand should face wherever the point is? So they're all in the positive x direction?

Yes , but aren't they equal and opposite? Wouldn't they just cancel each other?

Homework Statement How do I find the direction of the magnetic field at the points? Homework Equations Ampere's B = (UI) / (2*pi*r)[/B]The Attempt at a Solution I know how the Right hand rule, where your thumb is the direction of the current and your hand wraps the direction of the field...

Okay I see, so I2's field lines are opposite of the loop direction, so they make a negative contribution to the sum. So it should be B = Uo (-I2 + I3) Making I3 = (B/Uo) + I2 And I3 would be 26.2 A, and since it is positive it would be going into the page, correct?

Using the right hand rule to check the direction of positive I, positive I points into the page, so the 12A is negative and, so I should add 12 instead of subtracting?

Homework Statement The value of the line integral around the closed path in the figure is 1.79×10−5 Tm . https://session.masteringphysics.com/problemAsset/1385538/4/32-22.jpg There is I2 and I3 inside the closed loop. I2 = 12 A What is I3? Homework Equations Amphere's Law: B = Uo I(enclosed)...

Oh I see. So dx should be rdθ instead for polar coordinates right, without it I can't integrate from pi/2 to 3pi/2?

I integrated from pi/2 to 3pi/2. Which came out to be just pi and that canceled out the pi from L = pi * r.

Homework Statement There is a thin rod with charge Q that has been bent into a semicircle with radius R. Find an expression for the electric potential at the center. Radius = R Charge = QHomework Equations V = ∫(k * dq)/rThe Attempt at a Solution dq = λdx λ = Q/L L = pi * r V = kQ / LR ∫ dx =...

Ah! I see. I plugging in everything I know, the accelerations got canceled resulting in S = 0.01399 m. This makes sense because electrons move faster than protons.

Yes they are both in the denominator. Using the formula a = qE/m , I know that E should be the same for both the proton and electron, so I can set them equal. I'm not sure if this is what you were asking but I did (ae * me ) / qe = (ap * mp ) / qp Then solving for ae I got ae = ap * 1756.31

Ah I see. I know that t would be the same for both because that's when they meet with each other. Would I solve for t and make them equal to each and solve for S? t2 = 2S / ae and t2 = 2(D-S) / ap And solving for S would be... S = D * ae / ap + ae Unless my my algebra is wrong.

I'm not really sure how distance D relates to the problem, but my guess would be if electron is traveling from distance 0 to 1.4 cm and the electron is traveling from 1.4 cm to 0, if looking at it on an x-axis... then the distance the electron travels is S, while the distance the proton travel...

Homework Statement Two parallel plates 1.40 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate. How far from the negative plate is the point at...