91Ó°ÊÓ

i) Length of solenoid$L=85.0\mathrm{cm}=85.0×{10}^{-2}\mathrm{m}$

ii) Number of turns N = 950

iii) Current in the wire i = 6.60 A

iv) Cross-sectional area of the solenoid$A=17.0c{m}^2=17.0×{10}^{-4}{m}^2$

From the given total number of turns of wire, we can find the number of turns per meter. Using this number of turns into the formula of magnetic energy density, we can find the energy density of the magnetic field inside the solenoid. As the magnetic field is uniform inside the solenoid, using this calculated magnetic energy density and the given cross-sectional area and length of the solenoid, we can find the energy stored in the solenoid.

Formula:

$$\begin{gathered} u_B=\frac{1}{2}{\mathrm{μ}}_0{\mathrm{n}}^2{\mathrm{i}}^2 \\ {\mathrm{U}}_{\mathrm{B}}={\mathrm{u}}_{\mathrm{B}}\mathrm{V} \end{gathered}$$

Theenergy density of the magnetic field inside the solenoidis given by

$u_B=\frac{1}{2}{\mathrm{μ}}_0{n}^2{i}^2$

Here, we have

$$\begin{gathered} n=\frac{N}{L}=\frac{950\mathrm{turns}}{85.0×{10}^{-2}\mathrm{m}}=1.118×{10}^3{\mathrm{m}}^{-1} \\ \mathrm{Therefore} \\ {\mathrm{u}}_{\mathrm{B}}=\frac{1}{2}\left(4\mathrm{π}×{10}^{-7}\frac{\mathrm{H}}{\mathrm{m}}\right){\left(1.118×{10}^3{\mathrm{m}}^{-1}\right)}^2{\left(6.60\mathrm{A}\right)}^2 \\ {\mathrm{u}}_{\mathrm{B}}=34.2\frac{\mathrm{J}}{{\mathrm{m}}^3} \end{gathered}$$

Therefore, the energy density of the magnetic field inside the solenoid is$34.2\mathrm{J}/{\mathrm{m}}^3$

The energy stored in the solenoid is given by

$$\begin{gathered} U_B=u_{B}V \\ V=AL=\left(17.0×{10}^{-4}{\mathrm{m}}^2\right)\left(85.0×{10}^{-2}\mathrm{m}\right) \\ V=1.45×{10}^{-3}{\mathrm{m}}^3 \\ {\mathrm{u}}_{\mathrm{B}}=\left(34.2\frac{\mathrm{J}}{{\mathrm{m}}^3}\right)\left(1.45×{10}^{-3}{\mathrm{m}}^3\right) \\ {\mathrm{u}}_{\mathrm{B}}=4.94×{10}^{-2}\mathrm{J} \end{gathered}$$

Therefore, the total energy stored in the magnetic field is $4.94×{10}^{-2}\mathrm{J}$.