How Does Moving an Iron Rod Affect Ammeter Readings in a Solenoid Circuit?

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SUMMARY

The discussion centers on the effects of moving an iron rod in a solenoid circuit on ammeter readings. As the rod approaches the solenoid, magnetic induction occurs, causing an increase in current as the rod enters the solenoid. The reading on the ammeter will rise while the rod is inside the solenoid and then decrease as the rod moves away. The problem highlights the assumption of an ideal solenoid with no resistance, which would lead to infinite current, necessitating the inclusion of a resistor in series for realistic analysis.

PREREQUISITES
  • Understanding of electromagnetic induction principles
  • Familiarity with solenoid circuits and their characteristics
  • Knowledge of ammeter functionality and circuit analysis
  • Basic grasp of the relationship between voltage, current, and inductance (V = L di/dt)
NEXT STEPS
  • Study Faraday's Law of Electromagnetic Induction
  • Learn about the behavior of inductors in AC and DC circuits
  • Explore the effects of resistance in series circuits with inductive loads
  • Investigate practical applications of solenoids in electrical engineering
USEFUL FOR

Students studying physics, electrical engineers, and anyone interested in the principles of electromagnetism and circuit design.

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Homework Statement



A solenoid with a large number of turns of wire is connected in series with an ammeter and a battery.

An iron rod is moved towards the solenoid at constant speed.

What will happen to the reading on the ammeter

(A) as the rod moves closer to the solenoid, before entering the solenoid;
(B) as the rod is in the solenoid;
(C) as the rod moves away from the solenoid, after emerging from the solenoid.


2. The attempt at a solution

I think that magnetic induction will occur. Thus, the rod will be a temporary magnet attracting the solenoid. This will result in more current being drawn as the rod enters and exits from the solenoid. Is that right?
 
Last edited:
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V = d/dt(Li). Usually it's V = L di/dt but in this case ...

BTW the problem is badly stated. An ideal solenoid has no resistance so, in series with an ideal ammeter, the current would just build up to infinity before the rod is even inserted.

Assume a resistor in series with the solenoid, and an ideal ammeter (zero internal resistance).
 

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