Loop and solenoid, find the current

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SUMMARY

The discussion centers on calculating the current in a solenoid based on the torque experienced by a loop placed within it. The loop has an area of 0.01 m² and carries a current of 20 amperes, resulting in a magnetic dipole moment of 8000 A·m². The torque on the loop is given as π × 10⁻⁴ Newton meters, leading to the calculation of the magnetic field inside the solenoid as 3.93 × 10⁻⁸ Teslas. Ultimately, the current in the solenoid is determined to be approximately 3.075 × 10⁻⁴ amperes.

PREREQUISITES
  • Understanding of magnetic dipole moment (μ = NIA)
  • Familiarity with torque in magnetic fields (τ = μ × B)
  • Knowledge of magnetic field calculations in solenoids (B = μ₀/4π ∫(IdL/r²))
  • Basic principles of electromagnetism and solenoid physics
NEXT STEPS
  • Study the derivation of the magnetic field inside a solenoid with varying turn densities
  • Learn about the Biot-Savart Law and its applications in magnetic field calculations
  • Explore the relationship between torque and magnetic dipole moment in different configurations
  • Investigate the effects of loop orientation on torque in magnetic fields
USEFUL FOR

Physics students, electrical engineers, and anyone involved in electromagnetism or solenoid design will benefit from this discussion.

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Homework Statement



a single loop is placed deep within a 4 meter long solenoid having a total number of turns equal to 40000. the loop has an area od 0.01m^2 and it carries a current of 20 ampere. the loop is oriented so that the torque on the loop is a maximum with a magnitude of pi*10^-4 Newton meters.

what is the current in the solenoid?

Homework Equations



magnetic dipole moment, mu = NIA where N is number of turns, I is current, A is area

torque, tau = mu X B where X indicates cross product, B is magnetic field

magnetic field, B = mu_0/4pi[integral(IdL/r^2)] where mu_0 is constant = 4pi*10^-7, dL is change in length, r is radius/distance, I is current

The Attempt at a Solution



mu = NIA
mu = (40000)(20)(0.01)
mu = 8000

tau = mu X B
pi*10^-4 = 8000sin(90) ---> max torque so theta = 90 degrees
pi*10^-4/8000 = B
B = 3.93*10^-8 Teslas

area of circle = pi(r^2)
sqrt[0.01/pi] = r
r = 0.056 m

B = mu_0/4pi[integral(IdL/r^2)]
3.93*10^-8 = (10^-7(4)I)/(0.056^2)
I = ((3.93*10^-8)(0.056^2))/(4*10^-7)
I = 3.075*10^-4 ampere

correct approach? correct answer?
 
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For the torque on the loop placed inside the solenoid, that loop only has one turn, so its magnetic dipole moment will just be IA = 0.2 A·(m^2).

Also, somewhere in your chapter, there ought to be an equation for the strength of the ((nearly) uniform) magnetic field inside a solenoid, involving the "turn density" of the windings and the current through the solenoid's wire. (You wouldn't be able to use the Biot-Savart Law on the solenoid, anyway: you can't get a radius for it because we are not given the cross-sectional area for the solenoid. In any case, it isn't needed...)
 
Last edited:

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