Okay, so I took the derivative and got:
120pi*cos(120pi*t)((N(coil)mu(0)N(solenoid)I(0)pi*r^2)/L)
Do I need to solve for t by knowing that cos(120pit) is a max when 120pi*t = 0?
Even then it would seem that I'd be left with two unknowns: the emf and I(0)...
(sorry about the...
Hmmm... Let's revive this one.
I got as far as this:
Flux through coil = (N[SIZE="1"]coil*Mu[SIZE="1"]0*N[SIZE="1"]solen*I(t)*Pi*r^2)/L
I can then input my function for I into that equation and plug it into Farraday's law...
First, I know this is an old thread... But can you expand on this?
How can you solve for E = B*L*v if you don't know v? And then, how could you solve for I?
I think the logic here makes sense to me, but I am not making all the connections.
Thanks for your help everyone... I was able to solve this using my three equations. It was just a matter of plugging them into each other and avoiding stupid mistakes (which I am prone to).
I1 = .163
I2 = .212
I3 = I4 = .0490
I am still stuck on this problem. I came up with the equations:
I3 = I4
I2 = I1 + I3
(e-I2R2) - (I1R1) = 0
(e-I2R2) - (I3R3 + I4R4) = 0
I'm not sure why you say I have three equations and three unknowns... isn't it four? Or maybe you eliminated one since I3 = I4. Honestly, my algebra...
Also, what is the best way to solve the group of equations? I tried using a matrix, but I did not get the correct answer. It has been a looong time since I have attempted one, so it is possible I made an error.
I completely follow you, until I get here:
Why does the voltage drop on the right have to the the same as the voltage drop on the right?
Is it because the wires are at a junction at the top (and therefore must be at the same voltage) and there is a junction at the bottom (so these must...
Homework Statement
\epsilonbatt = 12V
Find the following currents.
The current I1 through the resistor of resistance R1 = 15.0 ohms.
The current I2 through the resistor of resistance R2 = 45.0 ohms.
The current I3 through the resistor of resistance R3 = 25.0 ohms.
The current I4 through the...
Homework Statement
An isolated 5.60 microfarad parallel-plate capacitor has 4.90 mC of charge. An external force changes the distance between the electrodes until the capacitance is 2.00 microfarads.Homework Equations
W = Fd
c = Q/VThe Attempt at a Solution
My first instinct was to solve to the...
Nothing was given about the pendulum length. I actually did the problem correctly except that I used 100/313 to find the period when I should have used 313/100.
The correct answer was g = 1.07m/s^2
I've repeated this result a few times... leading me to believe that possibly my pendulum calculation is incorrect because the mass of the wire has not been taken into account.
The length L for the pendulum is the length to the center of mass, perhaps that length isn't 2d? Is there another...