Answer:
a) 34.17J/m^3
b) 0.0468 J
Explanation:
a) Calculate the energy density of the magnetic field inside the solenoid as follows:
[tex]u_{B} &=\frac{U}{A l} \\ &=\frac{L I^{2}}{2 A l} \quad\left(U=\frac{1}{2} L I^{2}\right) \\ &=\frac{\mu_{0} N^{2} A I^{2}}{2 A l^{2}} \quad\left(L=\frac{\mu_{0} N^{2} A}{l}\right) \\ &=\frac{\mu_{0} N^{2} I^{2}}{2 l^{2}}[/tex]
Substitute the corresponding values in the above equation.
[tex]u_{B} &=\frac{4 \pi\left(10^{-7}\right)(1260)^{2}(8.78)^{2}}{2(0.693)^{2}} \\
&=34.17 \mathrm{~J} / \mathrm{m}^{3}[/tex]
b) Calculate the total energy stored in the solenoid as follows:
[tex]U &=u_{B} A l \\ &=(34.17)\left(24 \times 10^{-4}\right)\left(69.3 \times 10^{-2}\right) \\ &=0.0468 \mathrm{~J}[/tex]
