Recent content by gramentz

The only one that would be dependent on time would be B_{0} because the loop is held fixed and the angle does not change.

I'm thinking I have to use the emf at time t=0 to find the initial B at time t=0.

EMF= -\frac{B_{0}exp(-at)A}{dt} ? Or take the derivative of BAcos\phi with respect to time and set that equal to EMF? (Thank you for your help thus far though)

I am not sure I understand because I don't see any values given in the problem for the dimensions of this loop. Do you mean A = \frac{\phi}{B}?

\intBdA...B(dot)dA = BAcos\theta

The magnetic flux is a way of counting the total magnetic field lines. A changing magnetic field (given by the B(t)) will produce the emf's given.

Here is how the question is stated exactly: A circular conducting loop is held fixed in a uniform magnetic field thatt varies in time according to B(t) = B0exp(-at) where t is in s, a is in s-1 and B is the field strength in T at t=0. At t=0, the emf induced in the loop is 0.0758 V. At t=...

It took me quite some time, but I finally understand what you wrote and how you arrived at your answers. Thanks for taking the time to help me out, I appreciate it.

Then, I believe the complete solution will equal sin(4x) + cos(4x) + 5/16 cos(wt). Again, I'm not positive, but I think I did everything correctly...

Ohh ok right, somewhere along the line I copied down the equation to be a "-" and not a "+" for the gen. solution. I had to look it up, but I believe that sin(4x) and cos(4x) are equivalent because of imaginary numbers "rotating" between x and y, and the shifts between +i and -i correlate to the...

I have the general solution to equal c1e4x + c2e-4x And the particular solution (so far) to be c2cos(ax+\beta) + c2sin(ax+\beta) Does this look like I'm on the right track?

When we assume the general solution of the form y= csin(w0t) + dcos(w0t)...is this because all of the derivatives will take such form? Since dy/dt = v in the original equation I was given, that implies that step 7 in the thumbnail could also be v(t)? Thank you for helping me with this.

I arrived at that first "solution" because I made the mistake of thinking it was in fact dy/dv. Thanks for clearing that up for me. I'm going to work on this new setup now..

No, this isn't homework. Yes, the external force that is applied is given by 10sin(wt) where w will vary. I am first examining what happens when w = 2 rad/sec. I've been working on this and came up with the following at 2 rad/ sec (don't know if it is correct) dv/ dt = 5sin(2t) - 16y...

Mass on vertical spring with force applied. dy/dt = v dv/dt = 5sin(wt) - 16y I am trying to figure out the velocity of this object at any given time when w= 2 rad/ sec. (Would ideally like to have it for any w and any t) Did I do this correctly? (In calc 2, haven't had diff eq yet)...