Solenoid Induction Homework: Bz & Flux Calculation

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SUMMARY

The discussion focuses on calculating the magnetic flux through a circular conducting loop placed inside a solenoid. The solenoid has a length of 20 cm, a radius of 6 cm, and 4500 turns, while the loop has 11 turns and a radius of 1 cm. The current in the solenoid varies with time, described by the equation I1(t) = 0.45 A + 0.14 (A/s) t. The magnetic flux through the loop at t = 3 s is calculated to be approximately 7.36e-5 Tm², using the Biot-Savart law and the magnetic flux equation.

PREREQUISITES
  • Understanding of Biot-Savart Law for magnetic field calculations
  • Knowledge of magnetic flux and its calculation using the formula ϕ = ∫BdA
  • Familiarity with the concept of solenoids and their magnetic fields
  • Basic trigonometry to handle angles in magnetic flux calculations
NEXT STEPS
  • Study the derivation and applications of the Biot-Savart Law in electromagnetism
  • Learn about the properties and calculations related to infinite solenoids
  • Explore the effects of varying current on magnetic fields and flux
  • Investigate the relationship between the number of turns in a coil and induced magnetic flux
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or related fields seeking to understand solenoid behavior and magnetic flux calculations.

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Homework Statement


A solenoid has length L = 20 cm, radius 6 cm, and N1 = 4500 turns; its axis coincides with the z-axis. A circular conducting loop containing N2 = 11 turns of radius a = 1 cm is centered inside the solenoid; the plane of the loop makes a 30° angle with respect to the z-axis.

The current in the windings of the solenoid is varying with time according to the expression
I1(t) = 0.45 A + 0.14 (A/s) t.


Homework Equations



Biot-Savart(Center of a solenoid): Bz = u0nI
Magnetic flux: ϕ = ∫BdA= NBAcosϑ

The Attempt at a Solution



(a) Calculate the magnitude of the magnetic flux Φm through the loop at t = 3 s. (Absolute value)

I1(3s) = 0.45 A + 0.14 (A/s) (3s) = 0.87A

Bz = 4*pi*10^(-7)*(4500/0.2)*0.87 = 0.02459867 T

ϕ = (0.02459867 )*pi*(.01^2)*11*cos(30) = 7.3618137e-5 Tm^2

I did something wrong and I am not quite sure where. Any ideas?

I think i understand the concept. Since there is a small loop inside the solenoid at angle ϑ, I need to find the magnetic field at the center due to the current in the solenoid. Then I just have to find the total flux through the total area of the small circular loop that is sitting in the middle of the solenoid.

The question does provide a hint: Approximate this finite solenoid by an equivalent infinite solenoid to find the magnetic field in the central region.

Does this mean I can use a portion of the number of turns over a small distance to find the magnetic field?
 
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Help?
 
Are you sure that the magnitude of the flux through the small loop depends upon the number of turns in that loop?
 
Hmmm. I am not entirely sure. The equation for magnetic flux calls for it, but I think that is for the solenoid itself.
 
I know that the magnetic field is uniform with in the solenoid near the center, and that would mean that the loop will see what ever its area is.
 
I think I just answered your question! Thanks it makes a lot more sense now.
 

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