An Air-Filled Toroidal Solenoid

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SUMMARY

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 when a current of 11.7 A flows through it. The calculations utilized the equations for magnetic field strength (B), magnetic flux (Φ_B), energy stored (U), and inductance (L). The error in the initial calculation was identified as a misconversion of the cross-sectional area from cm² to m².

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Homework Statement


179190.jpg

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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