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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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.**

...

U=(1/2)*(I^2)*L

edited Mar 18, 2012 by Aaron Lee