91Ó°ÊÓ

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

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

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.

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

The magnetic field is given by

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

The self-inductance is given by

$$\begin{gathered} L=\frac{{\mathrm{μ}}_0{N}^{2}A}{2\mathrm{π}r} \\ L=\frac{4\mathrm{π}×{10}^{-7}×{300}^2×4×{10}^{-4}}{2\mathrm{π}×0.12} \\ L=6×{10}^{-5}H \end{gathered}$$

The energy stored is calculated by

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

The energy density is given by

$$\begin{gathered} u=\frac{B^2}{2{\mathrm{μ}}_0} \\ u=\frac{(2.5×{10}^{-3}{)}^2}{2×4\mathrm{π}×{10}^{-7}} \\ u=2.49J/m^3 \end{gathered}$$

The energy density is also given by

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

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