Answer:
[tex]65.6\ \text{J/m}^3[/tex]
[tex]0.11\ \text{J}[/tex]
Explanation:
B = Magnetic field = [tex]\mu_0 \dfrac{N}{l}I[/tex]
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]
N = Number of turns = 1270
[tex]l[/tex] = Length of solenoid = 93.9 cm = 0.939 m
[tex]I[/tex] = Current = 7.8 A
A = Area of solenoid = [tex]17.3\ \text{cm}^2[/tex]
Energy density of a solenoid is given by
[tex]u_m=\dfrac{B^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{(\mu_0 \dfrac{N}{l}I)^2}{2\mu_0}\\\Rightarrow u_m=\dfrac{\mu_0N^2I^2}{2l^2}\\\Rightarrow u_m=\dfrac{4\pi\times 10^{-7}\times 1230^2\times 7.8^2}{2\times 0.939^2}\\\Rightarrow u_m=65.6\ \text{J/m}^3[/tex]
The energy density of the magnetic field inside the solenoid is [tex]65.6\ \text{J/m}^3[/tex]
Energy is given by
[tex]U_m=u_mAl\\\Rightarrow U_m=65.6\times 17.3\times 10^{-4}\times 0.939\\\Rightarrow U_m=0.11\ \text{J}[/tex]
The total energy in joules stored in the magnetic field is [tex]0.11\ \text{J}[/tex].
