Calculate the energy associated with the magnetic field of a 201-turn solenoid in which a current of 1.29 A produces a magnetic flux of 3.76 *10-4 T · m^2 in each turn.

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Calculate the energy associated with the magnetic field of a 201-turn solenoid in which a current of 1.29 A produces a magnetic flux of 3.76 image 10-4 T · m2 in each turn.
 mJ

asked Mar 6, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

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The energy assoicated with a magnetic field is given by:

U =(1/2) L*I^2    (*where U=energy, I=current, L=inductance)

Since we know the current, we just need to find the inductance to solve for U:

Because we are given the magnetic flux we can use the following relation:

Magnetic flux = L*I  (*note, remember to multiply magnetic flux by the total number of turns to get the total magnetic flux)

Using this equation we can solve for L, and then plug back in the first equation to find U.

 

answered Mar 14, 2012 by kirby ~Expert~ (3,020 points)
edited Apr 14, 2012 by kirby
L=(N[flux]/I)
U=(1/2)*(I^2)*L
Comment by aaron lee  is correct
Answer has been corrected thanks

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