Question

The magnetic field in the interstellar space of our galaxy has a magnitude of about $\mathit{B}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{T}$. How much energy is stored in this field in a cube $\mathit{l}\mathbf{=}\mathbf{10}\mathbf{light}\mathbf{‐}\mathbf{years}$on edge? (For scale, note that the nearest star is 3.4light-yearsdistant and the radius of the galaxy is about 8 10⁴light-years. )

Short answer

The energy stored in the field is$U=3\times {10}^{36}\mathrm{J}$

Step-by-step solution

Given

$$\begin{gathered} B={10}^{-10}\mathrm{T} \\ l=10\mathrm{light}‐\mathrm{years} \end{gathered}$$

Understanding the concept

First we convert one light year to meter per second, after that, by using the relationship between energy density and magnetic field we calculate the energy stored in the field.

Formula:

$u_B=\frac{B^2}{2{\mu }_0}$

Calculate the energy stored in the field

According to the formula of energy density which is energy per unit volume if we multiply the volume of cube than we get energy stored in that field.

So first find the volume of the cube.

$$\begin{gathered} 1\mathrm{light}‐\mathrm{years}=3\times {10}^8\times 365\times 24\times 60\times 60 \\ 1\mathrm{light}‐\mathrm{years}=9.5\times {10}^{15}\mathrm{m}/\mathrm{s} \end{gathered}$$

The edge of the cube in m/s is

$$\begin{gathered} l=10\times 9.5\times {10}^{15} \\ l=9.5\times {10}^{16} \end{gathered}$$

So, the volume of the cube as

$$\begin{gathered} V=l^3 \\ V={\left(9.5\times {10}^{16}\right)}^3 \\ V=8.5\times {10}^{50} \end{gathered}$$

Now formula of energy density as
$$\begin{gathered} u_B=\frac{B^2}{2{\mu }_0} \\ U=\frac{B^2}{2{\mu }_0}V \end{gathered}$$

So, by substituting the value we can get,

$$\begin{gathered} U=u_B\times V \\ U=\frac{{\left({10}^{-10}\right)}^2}{2\times 4\mathrm{\pi }\times {10}^{-7}}\times 8.5\times {10}^{50} \\ U=\frac{{10}^{-20}}{2.5\times {10}^{-6}}\times 8.5\times {10}^{50} \\ U=3\times {10}^{36}\mathrm{J} \end{gathered}$$

Hence energy stored is $3\times {10}^{36}\mathrm{J}$