anonymous
  • anonymous
A very long cylindrical solenoid has a radius of 0.50m and 1000 windings per meter along its length. A circular conducting loop of radius 1.0m encircles the solenoid with the long central axis of the solenoid passing through the center of the loop, and with the area vector of the loop parallel to the solenoid axis. The solenoid initially carries a steady current I, but the current is then reduced to 0 during a 0.1s time interval. If the average emf induced in the loop during that interval is 0.1V, what was the initial current magnitude?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I was able to determine that\[\large B = \mu_0I \frac{N}{L}=\mu_0 In\]and that\[\large emf=\frac{FL}{q}=\frac{qvBL}{q}=vBL\]But I wasn't sure what to do with that once I got that.
IrishBoy123
  • IrishBoy123
connect your first statement, ie \(\large B = \mu_0I \frac{N}{L}=\mu_0 In\) to the answer via Faradays Law: \(\large \mathcal E = - \dot \Phi\) where \(\Phi = B A\) |dw:1446405632709:dw| the trick might be to get your head around the areas first. the loop is outside the solenoid. maybe do it first assuming that the loop and the solenoid have the same radius? then we take it from there...? this link shows some of the fundamentals, it may or may not help.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I don't see a link >_< But if it's hyperphysics, that's my go-to to find information :P
IrishBoy123
  • IrishBoy123
https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction i agree, hyper is really good too, great drawings etc; but you can't cut & paste latex :(. still a great free resource.
IrishBoy123
  • IrishBoy123
OMG!! emojis !
anonymous
  • anonymous
Irishboy can you help me
anonymous
  • anonymous
Yes please help me to!!!!😥
anonymous
  • anonymous
Okay, so then I got that \[\huge \mathcal E=\frac{-\mu_0 (\Delta I) n \pi r^2}{\Delta t}\]But what do I do with that? Which radius should I plug in? During the time interval 0.1s, the I was reduced to 0 and the emf induced was 0.1V. How could I find the initial I before it was reduced? Which radius would I plug in? @IrishBoy123
anonymous
  • anonymous
Hmm, okay so\[\huge 0.1V=\frac{-\mu_0I_0N \pi r^2}{0.1s(L)}\]Right? But then what's L or r?
anonymous
  • anonymous
N = 1000
anonymous
  • anonymous
Oh wait, it sayd that the solenoid contains 1000 windings PER METER, so I would plug n=1000, not N=1000. That gets rid of the L, no?
IrishBoy123
  • IrishBoy123
length disappears we need to find a way to distinguish between these situations. |dw:1446412134101:dw|
IrishBoy123
  • IrishBoy123
gtg but i will revert....
anonymous
  • anonymous
@IrishBoy123 Ok!
IrishBoy123
  • IrishBoy123
soz @CShrix , i was supposed to get back on this ..... and i suspect that you have already moved on, but here goes anyway 💥 because the solenoid and the coil have different areas, we have to be careful about how we match the equations we know that the B field from the solenoid is \(\large B = \mu_0 In\) the flux that we are talking about therefore, when we mention that Faraday's Law for the induced emf in the coil is \( \mathcal{E} = - \dot { \Phi } \), is \(\Phi = BA_s = \pi r_s^2 B = \pi r_s^2 \mu_0 In\) |dw:1446587553880:dw| so we have \(\large \mathcal{E} = -\dot{(\pi r_s^2 \mu_0 In)} \\ \large = -(\pi r_s^2 \mu_0 n)\dot I = -(\pi r_s^2 \mu_0 n)\frac{\Delta I }{\Delta t}\) so \(\large \Delta I = -\dfrac{ \mathcal{E} \Delta t} {(\pi r_s^2 \mu_0 n)} = -\dfrac{(0.1)(0.1)}{\pi (0.5^2)(4 \pi .10^{-7}) (1000)}\) that looks v similar to what you posted above : \[\large 0.1V=\frac{-\mu_0I_0N \pi r^2}{0.1s(L)}\] it gives: http://www.wolframalpha.com/input/?i=-%5Cdfrac%7B%280.1%29%280.1%29%7D%7B%5Cpi+%280.5%5E2%29%284+%5Cpi+*10%5E%7B-7%7D%29+%281000%29%7D hopefully close....🏁
anonymous
  • anonymous
@IrishBoy123 Interesting.. I definitely didn't go through the derivation that you did.. I used my equation that you noted here again to find my final solution.
IrishBoy123
  • IrishBoy123
is it the right answer?
anonymous
  • anonymous
I'm not sure! We turn in the assignments on paper and we usually don't get it back until weeks later.. I'll let you know once I get it back!
anonymous
  • anonymous
@IrishBoy123
IrishBoy123
  • IrishBoy123
do that! and good luck 😊

Looking for something else?

Not the answer you are looking for? Search for more explanations.