Question

In a proton accelerator used in elementary particle physics experiments, the trajectories of protons are controlled by bending magnets that produce a magnetic field of$\mathbf{4}\mathbf{.}\mathbf{8}\mathit{T}$. What is the magnetic-field energy in a$\mathbf{10}\mathit{c}{\mathit{m}}^{\mathbf{2}}$volume of space wherelocalid="1664180224552" $\mathit{B}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{8}\mathit{T}$?

Short answer

The energy is$91.7J$.

Step-by-step solution

The formulas involved in the problem

The energy density is given by$\mathit{u}\mathbf{=}\frac{{\mathbf{B}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{\mu }}_{\mathbf{0}}}$or the energy density is energy per volume.

Use the formulas to calculate energy

Given that$B=4.8T$and$V=10c{m}^3$. The energy density is given by

$$\begin{gathered} u=\frac{{\mathrm{B}}^2}{2{\mathrm{\mu }}_0} \\ u=\frac{4.8^2}{2\times 4\pi \times {10}^{-7}} \\ u=9.167\times {10}^{6}J/m^3 \end{gathered}$$

Since, the energy density is the energy divided by the volume, multiply energy density with the volume to find the energy.

$$\begin{gathered} U=uV \\ U=9.167\times {10}^6\times 10\times {10}^{-6} \\ U=91.7J \end{gathered}$$

So, the energy is $91.7J$.