Current in wire loop inside a solenoid

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SUMMARY

The discussion focuses on calculating the current in a wire loop placed inside a solenoid with specific parameters: a length of 2.1m, a diameter of 13.0cm, and 2200 turns of wire, with a current increase rate of 1.0 kA/s. Participants emphasize the importance of using Faraday's Law and the relationship between the solenoid's current and the magnetic field it produces, expressed as B(solenoid) = μ₀*n*I. The conversation also addresses the implications of a larger loop (19.5cm diameter) surrounding the solenoid, clarifying that the magnetic flux still passes through the loop despite the negligible external magnetic field.

PREREQUISITES
  • Understanding of Faraday's Law of Electromagnetic Induction
  • Knowledge of solenoid magnetic fields and their equations
  • Familiarity with the concept of magnetic flux
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the derivation of Faraday's Law and its applications in electromagnetic systems
  • Learn about solenoid magnetic field calculations, specifically B = μ₀*n*I
  • Explore the concept of magnetic flux and its relevance in larger loops surrounding solenoids
  • Investigate the effects of changing current on magnetic fields over time
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as engineers working with electromagnetic systems and applications involving solenoids and magnetic fields.

gills
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Homework Statement


A 2.1m -long solenoid is 13.0cm in diameter and consists of 2200 turns of wire. The current in the solenoid is increasing at the rate of 1.0 kA/s .


The Attempt at a Solution



Any help to point me in the right direction would be great.

I'm thinking that i need to find the magnetic field inside that of a larger solenoid. Then i somehow manage to put that into faradays law equation.
 
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That is exactly what you have to do. Good thinking.

HINT: How can you relate the current in the solenoid to the B-Field produced?
 
G01 said:
That is exactly what you have to do. Good thinking.

HINT: How can you relate the current in the solenoid to the B-Field produced?

how do i treat the current since it is given per unit time? That is throwing me off.

B(solenoid) = \mu_{0}*n*I ---> ??
 
ok this is what I've got

dB/dt = \mu_{0}*n*\frac{dI}{dt}
 
gills said:
ok this is what I've got

dB/dt = \mu_{0}*n*\frac{dI}{dt}

ok, got the first part.

What about when there's a loop that is larger than the solenoid? Would the current in that loop be zero since the magnetic field outside of solenoid is almost neglible?

The diameter of the larger loop is 19.5cm
 
gills said:
ok, got the first part.

What about when there's a loop that is larger than the solenoid? Would the current in that loop be zero since the magnetic field outside of solenoid is almost neglible?

The diameter of the larger loop is 19.5cm

ok, nevermind on that one. Would i just be changing the area of the loop from the equation i derived in the first problem or is there more to it?
 
gills said:
ok, nevermind on that one. Would i just be changing the area of the loop from the equation i derived in the first problem or is there more to it?

Yes, the only difference when the loop is bigger than the solenoid is the area of the loop. (The magnetic flux still goes through the loop if it's bigger than the solenoid.)

Keep working at it, you seem to be making some nice progress on this problem. Keep it up!:approve: If you have any more questions I'll answer them as soon as I can.
 

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