Question

An air-filled toroidal solenoid has a mean radius of$\mathbf{15}\mathit{c}\mathit{m}$and a cross-sectional area of$\mathbf{5}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. When the current is$\mathbf{12}\mathit{A}$, the energy stored is$\mathbf{0}\mathbf{.}\mathbf{39}\mathit{J}$. How many turns does the winding have?

Short answer

The number of turns is$2580$.

Step-by-step solution

The formulas used in the problem

The stored energy in the toroid solenoid is given by$\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{L}{\mathit{I}}^{\mathbf{2}}$.

The self-inductance in a toroid is given by$\mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{N}}^{\mathbf{2}}\mathbf{A}}{\mathbf{2}\mathbf{\pi }\mathbf{r}}$.

Use the two equations to find the number of turns

Given that$r=15cm,A=5c{m}^2,U=0.39J$and$I=12A$.

Using the two equations, the number of turns can be given by

$$\begin{gathered} N=\frac{2\pi rL}{{\mu }_{0}A} \\ N=\frac{2\pi \times 0.15\times 5.417\times {10}^{-3}}{4\pi \times {10}^{-7}\times 5\times {10}^{-4}} \\ N=2580 \end{gathered}$$

So, the number of turns is$2580$.