Question

An air-filled toroidal solenoid has $\mathbf{300}$turns of wire, a mean radius of $\mathbf{12}\mathit{c}\mathit{m}$, and a cross-sectional area of $\mathbf{4}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. If the current is $\mathbf{5}\mathbf{}\mathit{A}$, calculate: (a) the magnetic field in the solenoid; (b) the self-inductance of the solenoid; (c) the energy stored in the magnetic field; (d) the energy density in the magnetic field. (e) Check your answer for part (d) by dividing your answer to part (c) by the volume of the solenoid.

Short answer

(a)$B=2.5\times {10}^{-3}T$

(b)$L=6\times {10}^{-5}H$

(c)$U=7.5\times {10}^{-4}J$

(d)$u=2.49J/m^3$

(e)$u=2.49J/m^3$

Step-by-step solution

The formulas used in the given problem

The magnetic field is given by $\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{N}\mathbf{I}}{\mathbf{2}\mathbf{\pi }\mathbf{}\mathbf{r}}$.

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

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

The energy density is simply energy divided by the volume of the solenoid.

Calculate the magnetic field

Given that$r=12cm,A=5c{m}^2,N=300\mathrm{and}I=5A.$

The magnetic field is given by

role="math" localid="1664190143190" $$\begin{gathered} B=\frac{{\mu }_{0}NI}{2\pi r} \\ B=\frac{4\pi \times {10}^{-7}\times 300\times 5}{2\pi \times 0.12} \\ B=2.5\times {10}^{-3}T \end{gathered}$$

Calculate the self-inductance and the energy stored

The self-inductance is given by

$$\begin{gathered} L=\frac{{\mu }_0{N}^{2}A}{2\pi r} \\ L=\frac{4\pi \times {10}^{-7}\times {300}^2\times 4\times {10}^{-4}}{2\pi \times 0.12} \\ L=6\times {10}^{-5}H \end{gathered}$$

The energy stored is calculated by

$$\begin{gathered} U=\frac{1}{2}L{I}^2 \\ U=\frac{1}{2}\times 6\times {10}^{-5}\times 5^2 \\ U=7.5\times {10}^{-1}J \end{gathered}$$

Calculate the energy density by two ways

The energy density is given by

$$\begin{gathered} u=\frac{B^2}{2{\mu }_0} \\ u=\frac{(2.5\times {10}^{-3}{)}^2}{2\times 4\pi \times {10}^{-7}} \\ u=2.49J/m^3 \end{gathered}$$

The energy density is also given by

$$\begin{gathered} \mathrm{u}=\frac{Energy}{Volume} \\ \mathrm{u}=\frac{U}{2\pi rA} \\ \mathrm{u}=\frac{7.5\times {10}^4}{2\pi \times 0.12\times 4\times {10}^{-4}} \\ \mathrm{u}=2.49J/m^3 \end{gathered}$$

So, by using both the formulas, we get the same value of energy density.