How Does Changing Current in One Solenoid Affect Another?

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SUMMARY

When the current in the first solenoid, which has a magnetic field of 1.5 T and a cross-sectional area of 0.30 m², is uniformly reduced to zero over 0.5 seconds, it induces a voltage in the second solenoid wound around it. Applying Faraday's law, the induced voltage can be calculated using the formula V = dΦ/dt, where Φ represents the magnetic flux. With the induced voltage determined, Ohm's law can be utilized to find the current flowing in the second solenoid, which has 150 turns and a resistance of 20 ohms.

PREREQUISITES
  • Understanding of Faraday's law of electromagnetic induction
  • Knowledge of Ohm's law and its application
  • Familiarity with magnetic flux calculations
  • Basic principles of solenoid operation
NEXT STEPS
  • Study the derivation and applications of Faraday's law in electromagnetic systems
  • Learn about calculating magnetic flux in solenoids
  • Explore the relationship between induced voltage and current in circuits
  • Investigate the effects of self-inductance in solenoid configurations
USEFUL FOR

Physics students, electrical engineers, and anyone interested in electromagnetic induction and solenoid behavior.

klalli02
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A solenoid has a field of 1.5 T when the current is flowing in it. Its cross-sectional area is 0.30m^2. Another solenoid with 150 turns and a resistance of 20 ohms is wound tightly around the original solenoid so that both solenoids have the same cross-sectional area. The current in the first solenoid is reduced to zero uniformly over 0.5s. What current flows in the second solenoid during this time? Can some one help me with this question I am completely lost!
 
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Just apply Faraday's law to find the induced voltage, then Ohm's law to find the current. You can ignore self-inductance (I think) because the length of the solenoid is unknown.
 
You would just use V=dF/dt, where F is the flux. This should be easy (uniform change means just divide change in flux by change in time). Once you have the voltage and the resistance, I hope you can find the current
 

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