303 NOT 301 sorry

Anyone making any progress on this one?

Ive gotten to Vind = -d/dt(MagFlux)

And I found MagFlux to be 20B(t)cos(thetaBN)

I'll keep working on it...

UPDATE:

Ive gotten a value for B(t) my angle seems kinda high...over 100deg...
[Edited on April 9, 2007 at 5:09 PM. Reason : ...]

4/9/2007 6:08:38 PM

could you state the problem?

4/9/2007 6:41:33 PM

I got around 70 degrees...

Are your signs correct?

[Edited on April 9, 2007 at 7:19 PM. Reason : ]

[Edited on April 9, 2007 at 7:20 PM. Reason : ]

4/9/2007 7:19:04 PM

i get

20cos(theta)(0.000607) = -0.0068exp(-10t) + C

Is C zero? Then you would just solve for theta. How would I find the possible values for uB?

14.2 anyone?

4/9/2007 7:24:24 PM

What happens to B as t->[infinity]? What then must happen to theta as t->[infinity]? This will give you a limiting condition that will allow you to solve for C.

Double check the integration that you used to find an equation for theta.

[Edited on April 9, 2007 at 7:49 PM. Reason : stupid character sets...]

4/9/2007 7:43:21 PM

So saying C goes to zero is incorrect?

Look at the answers to 14.1 to solve 14.2.

4/9/2007 7:45:54 PM

Yes, the constant of integration is zero. The statement in the problem about B approaching zero as t approaches infinity gives you the information you need to establish a boundary condition that allows you to solve for C.

4/9/2007 7:51:57 PM

I'm still getting theta = 109 deg...I get 70 if it take off my negative sign...but when you integrate e to a - power you will get a minus.

Also any ideas on part B? I must be dense tonight XD

4/9/2007 8:40:51 PM

I am so lost right now where did the cosine come from. I guess i didn't do the problem correctly at all.

I am absolutly clueless though about 14-2. Could anyone please explain how to do this one.

4/9/2007 10:05:30 PM

14.2 is kinda daunting till you look at 14.1...

since Delta DOT J = -partial(p) / partial(t) and J = sigmaE and Delta DOT E = p / e

You can now say that -partial(p) / partial(t) = sigma * p(t) / e

solve for p(t)


I know its not a delta...its a triangle thingy
_____

the cos comes from the dot product of the normal vector to the wire and the unit vector of the B field (i think). since A DOT B = Amag * Bmag * cos(theta).

Im still confused on 13-10 part b...

[Edited on April 9, 2007 at 10:24 PM. Reason : .,..]

4/9/2007 10:22:28 PM

All the negative signs should go away in the end since vind = - d/dt etc......

You will end up with something like:

cos(thetaBN) = [(-#)*e^(-#t)] / [(-#)*B(t)] = +#

thetaBN = cos^-1 (+#)

[Edited on April 9, 2007 at 10:39 PM. Reason : ]

4/9/2007 10:36:42 PM

i see...

4/9/2007 10:40:44 PM

"its a triangle thingy" = nabla

4/9/2007 11:02:08 PM

for 14-2 does the dimensions of the thing not even matter

4/10/2007 12:05:37 AM

awww come on, post the problems.

4/10/2007 12:12:41 AM

for 13-10 b

here it is

Vinduced = -d/dt( integral(B*dS)) B=B*Ub dS= dS*Un

so when you put B and dS back into the Vinduced equations you have a dot product
pull out the scalars, the dot product left Ub*Un is mag of both times cos(theta) between them
pulll that out, all thats left is integral of dS = area of square

then plug in t=0.05 value and give then find cos(theta) then theta

4/10/2007 12:18:14 AM

HUR: I think you are correct...

4/10/2007 12:23:31 AM

Alexander must be slipping or I am missing something. He solves for p(t) in 14-1 part b, so all we have to do is plug in p0, sigma, and epsilon AND he gives the solution to the homework problem? What's that about?

4/10/2007 1:02:29 AM

^ I wondered about that too...

4/10/2007 10:19:23 AM