Magnetic flux through smaller coil within solenoid

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SUMMARY

The discussion centers on calculating the magnetic flux through a smaller coil located within a solenoid and determining the mutual inductance between the two coils. The solenoid has 235 turns, a length of 27 cm, and carries a current of 15 A, resulting in a magnetic field strength (B) of 0.01641 T. The smaller coil, with 50 turns and a diameter of 1.0 cm, requires the application of the formula for magnetic flux (flux = B x A) to find the total flux through it. The correct approach involves multiplying the magnetic field by the area of the smaller coil, taking into account its multiple turns.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically solenoids and coils.
  • Familiarity with the formula for magnetic field strength (B = μ₀ * n * I).
  • Knowledge of calculating magnetic flux (flux = B x A).
  • Basic grasp of mutual inductance concepts.
NEXT STEPS
  • Calculate the magnetic flux through the smaller coil using the formula flux = B x A with the correct area for 50 turns.
  • Explore the concept of mutual inductance and its calculation between two coils.
  • Investigate the effects of varying current in the solenoid on the magnetic flux in the smaller coil.
  • Learn about the application of Ampere's Law in calculating magnetic fields in solenoids.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone studying electromagnetic theory, particularly in the context of solenoids and inductive components.

jdubt
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Homework Statement


A solenoid has 235 turns of wire wrapped along its length of 27 cm. The diameter of the solenoid is 1.5 cm. In the middle of the solenoid is a smaller coil of diameter 1.0 cm with 50 turns of wire along its 2.0 cm length. The two coils are coaxial. A current of 15 A is in the larger solenoid with no current in the smaller coil.
(a) Determine the magnetic flux through the smaller coil.
(b) Determine the mutual inductance of the pair of coils.

Homework Equations


B= mu_not *n*I
flux =B x A


The Attempt at a Solution


I've tried several times with no success. My problem is I don't know how to relate the solenoid to the smaller coil.

B= 4* pi *10^-7*(235/.27)*15 = .01641 T

I believe this is correct but I don't know where to go from here.

flux = (.01641 T)* pi * (.0075)^2

This could be an intermediary step in getting the flux of the smaller coil, but I don't know what to afterwards.

flux = (.01641 T)* pi * (.005)^2

This is incorrect.

Any help would be much appreciated.
 
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I agree with your 0.01641 T calculation.

Hmmm ... your flux expression seems right for a single-loop coil, but they give 50 turns for the smaller coil.
 

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