EMI Quickies: Get Answers to Your Questions

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The discussion revolves around electromagnetic induction and circuit behavior, particularly focusing on current changes when a switch is closed or opened. Participants explore the relationship between current, time, and resistance in a circuit, emphasizing that inductance is influenced by the geometry of the solenoid. Questions arise about the behavior of electrons and the production of heat during these processes, with clarification that electron movement does not differ significantly between stationary and moving rods. The conversation also touches on the mathematical relationships governing current flow and the implications of resistance in the circuit. Overall, the thread seeks to clarify complex concepts in electromagnetism and circuit dynamics.
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#1
No idea.

#2
Please someone explain me the question. From what I have understood, I think it’s because on connecting wires, it will form a closed loop and will induce an opposing EMF which will induce current in circuit and which will provide an opposing torque to the motion.

#3
Yes, start at instance battery is connected and continue to rise till infinity as current in first will also me maximum at infinity and will keep on increasing till then

#4
Same reason as third, current will be 0 at infinity

#5
Doesn’t inductance only depend on just geometry of solenoid? Then why will it change for ends?

#6
I guess at center as magnetic field lines are denser at center.

#7
The rod will magnetize so field and flux will increase, but answer is (a), (b), (c)

#8
Answer says that (d) is wrong, why? While electron will shift at B, they will collide and produce heat, right?
 

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1. How does the current change with time after the switch is closed? What is the time derivative of the current at t=0? And how much is the change of the current when the switch is opened? How long can it last to open a switch?

ehild
 
For closing: i = io (1 - e-t/τ)
so di/dt = ioe-t/τ / τ
at t=0, i = io

For opening, i = io e-t/τ
di/dt = -io e-t/τ / τ
di/dt = -io / et/τ τ

As 1 > e-x > 0 for x>0

di/dt in opening is greater than one ... right?
 
What is Io when the switch is closed? How is it related to the battery and the resistance in the circuit? There is always some resistance: the inner resistance of the battery, the resistance of the wires, the resistance of the coil (it is made of wire which has some resistance)

Can current flow at all when the switch is open? The equation you wrote refers to the case when the battery is short-circuited at t=0, at the current is io at that instant. It is not breaking the circuit.

ehild
 
8. no reason it will not heat up, unless it has no resistance or there is only 1 correct answer, in which case end A will become positively charged answer (C) (because of Lorenz force).
 
cupid.callin said:
#8
Answer says that (d) is wrong, why? While electron will shift at B, they will collide and produce heat, right?
There is no more movement of electrons than there is for a stationary rod.
 
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