Solving Inductance Problem: Flux, Capacitance & Voltage

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SUMMARY

The discussion focuses on calculating the voltage across a capacitor connected to a 200-turn coil subjected to a magnetic field of 55 μT. The coil, with a diameter of 4.0 cm and a resistance of 2.0 Ω, is rotated 180° after being discharged. The key equations utilized include the magnetic flux equation \\Phi = ABcos\\theta and the voltage across the capacitor Vc = Q/C. The final voltage across the capacitor is determined to be 12V, despite initial calculations suggesting otherwise.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with the equations of magnetic flux and capacitance
  • Knowledge of coil resistance and its impact on current
  • Basic proficiency in calculus for differentiating magnetic flux
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  • Study the principles of Faraday's Law of Electromagnetic Induction
  • Learn about the relationship between inductance and capacitance in RLC circuits
  • Explore the application of the formula Vc = Q/C in capacitor discharge scenarios
  • Investigate the effects of coil orientation on induced voltage
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Students in physics or electrical engineering, educators teaching electromagnetic theory, and anyone involved in practical applications of inductance and capacitance in circuits.

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inductance, flux, capacitance
1. Homework Statement
The magnetic field at one place on the earth’s surface is 55 μT in strength and tilted 60°
down from horizontal. A 200-turn coil having a diameter of 4.0 cm and a resistance of 2.0 Ω
is connected to a 1.0 μF capacitor rather than to a current meter. The coil is held in a
horizontal plane and the capacitor is discharged. The coil is then quickly rotated 180° so that
the side that had been facing up is now facing down. Afterward, what is the voltage across
the capacitor? [Hint: use I=dq/dt


2. Homework Equations
LaTeX Code: \\Phi =ABcosLaTeX Code: \\theta
Vc=Q/C
Vl=-L(di/dt)
Vc=-Vl
I=LaTeX Code: \\Phi /L
LaTeX Code: \\Phi =NLaTeX Code: \\Phi per coil

3. The Attempt at a Solution
ok wat i did was find flux per coil=LaTeX Code: \\pi r2BcosLaTeX Code: \\theta =3.45x10-8
then flux=3.45x10-8*200=6.9x10-6=L(di/dt)=Q/C...
but that can't be right coz i know the answer is 12V... can someone please help me, I've been on this for 4 or hours and its 3am almost... due in tomoz... please someone help...
 
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any ideas??
 
please I am getting disparate here... well emf=|d[tex]\Phi[/tex]/dt|=L*di/dt=6.91x10-6... does not sound right at all!
emf=IR=>I=emf/R=... oh id don't know please!
 

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