Recent content by physics24sam

Sorry about that... a) there are 1.1x10^4 turns in the solenoid b) In regards to the .35m, I'm not sure. That is how it is worded in the problem. All it says is .35m inside which I assumed to mean .35m radially inward. c) My professor posted an answer key, so I do know the right answer. He...

Homework Statement The current in a solenoid (you may treat it as a long solenoid of length 2.0 m, turns, and radius 0.5 m) is decreasing at a rate of 1.7 A/s. What is the EMF at a point .35m inside the solenoid? Homework Equations B=N(mu_0*I)/(2a) where B=magnetic field, N=turns per...

Okay, so I got E_net=4kQ/∏a^2 as my final magnitude and it is pointing in the +x-direction.

I understand that the dE_x vectors both point in the +x-direction, but if I put in the integral of cosΘdΘ over the interval ∏/2 to ∏ on my calculator it is giving me the answer of -1. The answer is positive one for an interval of 0 to ∏/2, but I've been looking at the second quadrant for the...

Isn't the answer for the integral going to be -1 either way?

I thought I was already taking that approach... My integration limits are ∏/2 to ∏. Wouldn't that only cover the second quadrant?

So I was right in thinking that my dL=adΘ! That makes me feel a little better... I have already decided that I was originally wrong and that λ=Q/.5∏a since the arc length of a quarter-circle is ∏r/2. dE=kdq/r^2 → dE=k(Q/.5∏a)(adΘ)/r^2 → dE=kQdΘ/.5∏r^2 To get the x-component of that do I...

I knew already that the y-components would cancel and I would only have to deal with x-components. Would that change my limits somehow? I know for a line dq=λdL. I assumed that λ=Q, but now I am rethinking that since λ is charge per length and Q is just charge... Would dq=(Q/.5∏a)dL →...

I didn't even think to use polar coordinates since we haven't yet in class, but it makes perfect sense. I substituted λ for Q because the charge of the quarter-circle was Q. If it is supposed to be λ in that equation then how will I eventually get rid of that variable?

Homework Statement A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the...