In order to do that, I'll need an integral containing the entire area of the loop. This just got way more complicated than I thought it would be. How am I supposed to apply this integration in this problem?
If I use i(3) then I'd receive the amount of current at the time t=3. If I use just the equation, I'd probably end up having to take the derivative with respect to time. Is that correct?
I'm not really aware of any magnetic field equations that would fit this problem without an R term... but...
Oh alright so I(3) = what I want for I in the equation of the magnetic field and from what you're saying I think I have the right B equation. This was my resource (shows how pi drops out).
http://faculty.wwu.edu/vawter/physicsnet/topics/MagneticField/MFLoops.html
Would r in B = ui/2r be the...
Homework Statement
A loop of wire with an area 1cm^2 is centered within a larger loop of area 1m^2. Apower supply is generating current through the larger loop described by the function:
I(t) = 40A * cos(2t/s * 2pi/5)
Calculate EMF at time t = 3sHomework Equations
Magnetic flux =...
Oh right, duh.
Alright so I was probably right in thinking it requires the F = qvBsin(theta). Where theta will probably be 16 degrees.
However, that q in that equation confuses me.
So this is what I got now:
F = BIL = mv^2/r = qvBsin(theta)
I = mv^2/(BLR) = qvsin(theta)/L
So I have a...
Yea the right hand rule is a bit confusing but I think I got it.
The force you are talking about is F1 = B2I1L where L is the length (6cm).
And since the current is equal and with this geometry, the forces are equal. As such, both their magnetic fields should be equal.
Also since it's...
Homework Statement
A telegraph consisted of 2 long straight wires carrying currents equal in magnitude but oppositely directed. Each was suspended by a pair of light-weight and inelastic threads that were about 6cm long. When current flowed through the wires, the magnetic force exerted between...
Ah duh.
So EA = Q/ε = ρv/ε
A cancels out v except leaves r/3 on the right.
E = ρr/3ε
Eq = F = ρqr/3ε
Plugging the numbers in gets me closer, but not quite there. Perhaps there's a typo in the homework thing. But now I finally understand and I found a similar problem in my book and it matches...
Ah I see. Here's my new work.
EA = E4pir^2 = Q/ε
so E = Qr/(4*pi*ε*R^3)
Q/(4piR^3)= charge density
E = ρr/ε
And if F = E*q
F = ρrq/ε = (2*10^-3)(.01)(-1.602 * 10^-19)/(8.85*10^-12) = -3.62*10^-13 Nt
This is also wrong, answer should be in a lower magnitude around 10^-17. I've...
Homework Statement
The charge density of a spherically symmetric and uniform distribution of charge is ρ = 2*10^-3 C/m^3. An electron is released one centimeter from the center. Find the magnitude and direction of the force onto the electron.Homework Equations
charge density = Q/V =...
Not a specific question, but I just need help understanding how units cancel out.
v = sqrt(force of tension / mass density)
Force of tension is in Newtons. Mass density is in kg/m^3
Nt/ (kg/m^3) = (kg*m/s^2)/(kg/m^3) =(I cross multiply here) (kg*m*m^3)/(s^2*kg)
kg cancels out...
Homework Statement
An object emitting a sound with a 400hz frequency is thrown 10m away and passes right next to you. At 10m away, you can barely hear the object and you have average hearing. When the object passes you, you observe a sound with a frequency of 435hz.
Find the loudness in dB...
I think I was just stuck thinking about oscillations. From what I remember, when y=0 it's at a max velocity and at the top or bottom of its amplitudes, it's at its max acceleration. Something like that.
Thanks though.