Induced current with two solenoids

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SUMMARY

The discussion focuses on calculating the induced current in an inner solenoid placed within an outer solenoid. The inner solenoid has a resistance of 11 ohms, 1000 turns, and a radius of 0.02 meters, while the outer solenoid has 18000 turns per meter and a current increasing at a rate of 0.7 Amps/sec. The correct induced current is determined using Faraday's Law of Induction and Ampere's Law, leading to a final answer of 1.81(10-3) amps. The discussion emphasizes understanding the derivation of equations rather than just applying them blindly.

PREREQUISITES
  • Understanding of Faraday's Law of Induction
  • Familiarity with Ampere's Law
  • Knowledge of solenoid properties and magnetic fields
  • Basic circuit theory, including Ohm's Law
NEXT STEPS
  • Study the derivation of Faraday's Law and its applications in electromagnetic induction
  • Explore the relationship between magnetic fields and current in solenoids
  • Learn about the concept of magnetic flux and its calculation in solenoids
  • Investigate the effects of resistance on induced currents in electrical circuits
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Students studying electromagnetism, physics educators, and anyone seeking to understand the principles of induced currents in solenoids.

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Homework Statement



so there's two solenoids on the same axis of symmetry, and the smaller one inside the larger one has a resistance of 11 ohms, 1000 turns, and radius .02 meters. the outer solenoid has n 18000 turns per meter, and had a current that is increasing at a constant rate with dI/dt=.7 Amps/sec. The inner solenoid is connected to an ammeter.

The Attempt at a Solution



how would you find the induced current in the inner solenoid? i have an equation i tried using:
Iind=[(μ0)(n(turns per meter))(Imax)(A)]/[(R)(Δt)]
made μ0=4pi(10-7), A=pi(0.022) R=11, and since dI/dt=0.7 A/s i just made change in time equal 1 sec and Imax equal 0.7 Amps. One thing I think that may be messed up is the value for n, in the problem were given how many exact turns it has which is N, and no length of the solenoid(s). But the equation i have uses n which is turns per meter. So i just put for n (18000).

Got 1.91(10-6) amps, but the correct answer is 1.81(10-3) amps...tried a few other things and I am just kind of lost. Can somebody help me understand this?
 
Last edited:
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Thanks for the help guys -_-
 
From Ampere's Law you can derive that the magnetic field B in a solenoid is:

B_{solenoid} = μ_{0} n I

Then, you can use Faraday's Law of Induction to find the electromotive force in the second solenoid:

ε = - \frac{\partial \Phi_{B}}{\partial t} = - μ_{0} n \frac{dI}{dt} A N,

where A is the area of the second solenoid and N is the number of loops(this is actually an integral but since B is approximately constant, then the flux is just the magnetic field times the area. Since the solenoid has many loops, the area is multiplied by N since you're calculating the flux through each loop.

This gives you ε, from which you can calculate the induced current since ε = I_{induced} R.

This should give you enough info to solve the problem.

Also, the negative sign indicates the induced current is in the opposite direction, but for your case you can just take the absolute value.
 
1. Compute the flux inside the larger coil as a function of applied current.
2. Use Farady's law to determine the emf generated in the smaller coil.
3. Use i = emf/R for the current of the smaller coil. (An ammeter shorts the smaller coil).

Don't start with an equation with delta t in it. You're using equations without understanding their origin. What's important is to understand what's going on, not pulling equations you "have" to hopefully solve a given problem.
 
unfortunately i have a teacher that just makes us watch him derive equations and then doesn't explain anything or how it worked so i get a little lost. but i got the answer now, thanks
 
seriously, "rude_man" is right. I don't get why some people on this site have to act so condescending about physics. @rude_man, i am new to this subject...i tried really hard to understand it myself but was still having trouble, so i came here. i don't need people like you giving me attitude like that. it happens way too much on this site. Thanks LoadedAnvils, you didnt care to criticize me and simply explained how the problem works and how to find the answer. now i understand it. much appreciated
 
No problem. Physics is a very hard subject (I had a hard time with it even though I liked the subject), and sometimes the only way to understand something is to be walked through it step by step.
 

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