At what rate must the current be to changed to produce 40V emf

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Homework Help Overview

The problem involves an inductor with a specified inductance and current, seeking to determine the rate of change of current required to produce a certain electromotive force (emf). The subject area is electromagnetism, specifically focusing on inductors and their behavior under changing currents.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between steady current and the rate of change of current, questioning how to interpret the given steady current in the context of the emf equation. There is exploration of the formula EMF = -L di/dt and attempts to isolate di/dt.

Discussion Status

Some participants have offered insights into the implications of a steady current on the calculation of di/dt, while others have acknowledged the importance of the negative sign in the formula. The discussion reflects a mix of interpretations regarding the relevance of the steady current value provided.

Contextual Notes

Participants are grappling with the implications of the steady current condition and its effect on the calculation of the rate of change of current. There is an acknowledgment of the formula's requirements and the potential confusion arising from the problem's wording.

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A 20 H inductor carries a steady current of 4.0 A. At what rate must the current be changed to produce a 40 V emf in the inductor?

I odn't get this one because it gives you all the information!
The formula is:
EMF = -L di/dt
they want u to find di/dt, but they already give you it has a steady current of 4.0A, so isn't di/dt just i, which is 4.0A?
Then the EMF would be 40, and they say L = 20H, so I'm stuck on what intermediate step I'm missing! thanks
 
Last edited:
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For a steady current, at any value,

[tex]\frac {dI} {dt} = 0[/tex]
 
So basically, I can disregard that they told me it had a steady current of 4.0A right? and then just use
di/dt = EMF/-L to find the di/dt?
 
thanks that was the trick! i did that the last time, but forgot the minus sign!
 

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