Answer:
(a) Energy density will be equal to [tex]57.31J/m^3[/tex]
(b) Total energy will be equal to 0.0718 J
Explanation:
It is given that length of solenoid l = 78.8 cm = 0.788 m
Cross sectional area [tex]A=15.9cm^2=15.9\times 10^{-4}m^2[/tex]
Number of turns of the wire N = 914
Current in the solenoid i = 8.25 A
Inductance of the wire is equal to [tex]L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H[/tex]
(b) Total energy stored in magnetic field [tex]U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J[/tex]
(a) Energy density will be equal to
[tex]U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3[/tex]
