20.0-mH inductor potential energy problem

AI Thread Summary
The discussion centers on calculating the magnetic potential energy stored in a 20.0-mH inductor when carrying a current of 0.2 A, with the correct answer being 4.0 * 10^-4 J. Participants confirm the use of the formula for energy stored in an inductor, E = (1/2)L*I^2, for both parts of the problem. For part b, the same formula is suggested to find the current needed to store 1.0 J of energy, leading to the equation 1 = (1/2)*0.020*I^2. There is a request for verification of the calculations, indicating some confusion about the results. The discussion emphasizes the application of inductance principles in solving energy-related problems.
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Inductance
You are given a 20.0-mH inductor by your laboratory instructor, who then asks you to place it in series with a power supply. Before instructing you to turn on the power supply, he quizzes you on your understanding of the behavior of solenoids.

a. How much magnetic potential energy is stored in the inductor when it carries a current of 0.2 A?
Possible answers: a) 1.3 * 10^-4 J
b) 2.7 * 10^-4 J
c) 4.0 * 10^-4 J
d) 5.1 * 10^-4 J

b. What should the current in the coil be in order that it contains 1.0 J of energy?
Possible answers: a) 10.0 A
b) 15.0 A
c) 20.0 A
d) 25.0 A

i think i got a) energy stored = (1/2)L*I^2
= .5 * .020 * .2^2
= .00040 J -> 4*10^-4 J

can anyone help me with letter b)?
 
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a.) looks good.

for b.) why not use the same equation as part A?
 
so b is 1=(1/2)*.020* I^2 then?
is the answer .01 A?
 
Check your math.
 
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