Question

An air-filled toroidal solenoid has a mean radius of 15.5 cm and a cross-sectional area of 5.00 cm^2 When the current is 12.5A, the energy stored is 0.395J

How many turns does the winding have?

Answer 1

Concepts and reason

The concept is used to this problem is to energy stored by the inductor.

Initially form the expression of the energy stored by the inductor the inductance of the system is determined. Form the definition of the inductance the number of the turns of the system is determined.

Fundamentals

The expression of the energy stored by the inductor is expresses as follows,

$E = \frac{1}{2}L{I^2}$

Here, $E$ is the energy steroid by the inductor, $L$ is the inductance and $I$ is the current.

The expression of the inductance of the system is expresses as follows,

$L = \frac{{{\mu _0}{N^2}A}}{{2\pi R}}$

Here, ${\mu _0}$ is the permeability, $N$ is the number of the turns, $A$ is the area and $R$ is the radius of the coil.

The expression of the energy stored by the inductor is expresses as follows,

$E = \frac{1}{2}L{I^2}$

Substitute $0.395{\rm{ J}}$ for $E$ and $12.5{\rm{ A}}$ for $I$ in the above expression of the energy stored by the inductor,

$\begin{array}{c}\\\left( {0.395{\rm{ J}}} \right) = \frac{1}{2}L{\left( {12.5{\rm{ A}}} \right)^2}\\\\L = \frac{{\left( 2 \right)\left( {0.395{\rm{ J}}} \right)}}{{{{\left( {12.5{\rm{ A}}} \right)}^2}}}\\\\ = 0.005056{\rm{ H}}\\\end{array}$

Calculate the number of the turns of the inductance.

Thee expression of the inductance of the system is expresses as follows,

$L = \frac{{{\mu _0}{N^2}A}}{{2\pi R}}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ H/m}}$ for ${\mu _0}$ , $0.005056{\rm{ H}}$ for $L$ , $5.00{\rm{ c}}{{\rm{m}}^2}$ for $A$ and $15.5{\rm{ cm}}$ for $R$ in the above expression of the inductance of the system,

$\begin{array}{c}\\\left( {0.005056{\rm{ H}}} \right) = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right){N^2}\left( {5.00{\rm{ c}}{{\rm{m}}^2}} \right)}}{{\left( {2\pi } \right)\left( {15.5{\rm{ cm}}} \right)}}\\\\{N^2} = \frac{{\left( {0.005056{\rm{ H}}} \right)\left( {2\pi } \right)\left( {15.5{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {5.00{\rm{ c}}{{\rm{m}}^2}} \right){{\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}^2}}}\\\end{array}$

Taken square roots both sides,

$\begin{array}{c}\\N = \sqrt {\frac{{\left( {0.005056{\rm{ H}}} \right)\left( {2\pi } \right)\left( {15.5{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ H/m}}} \right)\left( {5.00{\rm{ c}}{{\rm{m}}^2}} \right){{\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right)}^2}}}} \\\\ = 2799{\rm{ turns}}\\\\ \approx {\rm{2800 turns}}\\\end{array}$

Ans:

The number of turns of the inductance is equal to $2800{\rm{ turns}}$ .