The current in a 20.0-mH inductor changes with time as I = 2.00t2 − 5.00t, where I is in amperes and t is in seconds.

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The current in a 20.0-mH inductor changes with time as I = 2.00t2 − 5.00t, where I is in amperes and t is in seconds.

(a) Find the magnitude of the induced emf at t = 1.00 s.
 mV

(b) Find the magnitude of the induced emf at t = 4.00 s.
 mV

(c) At what time is the emf zero?
 s
asked Mar 6, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
retagged Mar 12, 2012 by Joey33
    

1 Answer

+2 votes
 
Best answer

Emf is given by the formula:

Emf = -L * dI/dt  (*where L=inductance, I=current, t=time)

L = 20 mH = 20*10^-3 H

I = 2.00t2 − 5.00t,

taking the derivative we get dI/dt

dI/dt = 4t -5

 

Part A, at t = 1 s,

dI/dt = 4(1) -5 = -1 A/s

and so Emf is:

Emf = - 20*10^-3 H * -1 A/s  = .02 V = 20 mV  

Part B is solved in the same way.

 

Part C,

the emf is zero when dI/dt = 0

so setting dI/dt = 0, we get

dI/dt = 4t -5 = 0,

so  dI/dt = 0, when t = 5/4 s

 

 

answered Mar 12, 2012 by Roar3215 ~Expert~ (1,318 points)
edited Mar 12, 2012 by Roar3215



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