Answer:
(a) Energy Density = 160.94 J/m³
(b) Energy Stored = 0.192 J
Explanation:
(a)
The energy density of the magnetic field inside the solenoid is given by the following formula:
[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]
where,
B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]
Therefore,
[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]
where,
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 1300
I = current = 8.15 A
L = length = 66.2 cm = 0.662 m
Therefore,
[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]
Energy Density = 160.94 J/m³
(b)
Energy Stored = (Energy Density)(Volume)
Energy Stored = (Energy Density)(Area)(L)
Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)
Energy Stored = 0.192 J
