Question

It is proposed to store$\mathbf{1}\mathit{k}\mathit{W}\mathbf{.}\mathit{h}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{J}$of electrical energy in a uniform magnetic field with magnitude$\mathbf{0}\mathbf{.}\mathbf{6}\mathit{T}$. (a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy? (b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube$\mathbf{40}\mathit{c}\mathit{m}$on a side, what magnetic field is required?

Short answer

a) $V=25.1m^3$

b)$B=11.9T$

Step-by-step solution

The formulas involved in the given problem

The energy density can be calculated by $\mathit{u}\mathbf{=}\frac{\mathbf{U}}{\mathbf{V}}$or $\mathit{u}\mathbf{=}\frac{{\mathbf{B}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}}$.

Equate both the equations and find the volume.

Calculate the volume

Given that$U=3.6\times {10}^{6}J,l=40cm$and$B=0.6T$. Using both the energy density equations

$$\begin{gathered} \frac{U}{V}=\frac{B^2}{2{\mu }_0} \\ V=\frac{2{\mu }_{0}U}{B^2} \\ V=\frac{2\times 4\pi \times {10}^{-7}\times 3.6\times {10}^6}{0.6^2} \\ V=25.1m^3 \end{gathered}$$

The side of the cube is$l=40cm$. The volume will be $0.064m^3$.

$$\begin{gathered} B=\sqrt{\frac{2{\mu }_{0}U}{V}} \\ B=\frac{2\times 4\pi \times {10}^{-7}\times 3.6\times {10}^6}{0.064} \\ B=11.9T \end{gathered}$$