An Air-Filled Toroidal Solenoid

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An air-filled toroidal solenoid with a mean radius of 14.5 cm and a cross-sectional area of 4.99 cm² requires a minimum of 287 turns to store at least 0.388 J of energy with a current of 11.7 A. The calculations involved the magnetic field equation and the energy formula, leading to the determination of the number of turns needed. An error was identified in the conversion of the area from cm² to m², which was corrected to achieve the final result. The correct conversion factor is 1 cm² = 0.0001 m². The solution confirms the importance of accurate unit conversions in physics calculations.
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Homework Statement


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An air-filled toroidal solenoid has a mean radius of 14.5 cm and a cross-sectional area of 4.99 cm^2 (see the figure). The current flowing through it is 11.7 A, and it is desired that the energy stored within the solenoid be at least 0.388 J.

What is the least number of turns that the winding must have?
Express your answer numerically, as a whole number, to three significant figures.

Homework Equations



B = \frac {\mu_0 N I}{2 \Pi r}

\Phi_B = \oint \vec{B} \cdot \vec{dA}

U = \frac {1}{2} L I^2

L = \frac {N \Phi_B}{i}

The Attempt at a Solution



Solved for N to get the Number of Turns.

N = \sqrt{\frac {4 (\Pi) U r}{\mu_0 I^2 A}}

N = \sqrt{\frac {(4) (\Pi) (0.388j) (0.145m)}{(4 (\Pi) (10^{-7}) \frac{wb}{Am}) (11.7A^2) (0.0499m^2)}}

N = 287 turns

Do I have some conversion wrong or did I miss something.
 
Last edited:
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SOLVED!

Found my error in my conversion of cm^2 to m^2.

I did my the correct conversion is \frac{4.99}{10000}
 
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