Answer:
(a) The current needed is 56.92 A
(b) The magnitude of the magnetic field inside the solenoid is 0.134 T
(c) The energy density inside the solenoid is 7.144 kJ/m³
Explanation:
Given;
energy stored in the magnetic field of solenoid, E = 8.1 J
number of turns of the solenoid, N = 620 turns
diameter of the solenoid, D = 6.6 cm = 0.066 m
radius of the solenoid, r = D/2 = 0.033 m
length of the solenoid, L = 33 cm = 0.33 m
Inductance of the solenoid is given as;
[tex]L= \frac{\mu_o N^2 A}{l}[/tex]
where;
A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
[tex]L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H[/tex]
(A). How much current needed
Energy stored in magnetic field of solenoid is given as;
[tex]E = \frac{1}{2} LI^2\\\\[/tex]
Where;
I is the current in the solenoid
[tex]E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A[/tex]
(B) The magnitude of the magnetic field inside the solenoid
B = μ₀nI
where;
n is number of turns per unit length
B = μ₀(N/L)I
B = (4π x 10⁻⁷)(620/0.33)(56.92)
B = 0.134 T
(C) The energy density (energy/volume) inside the solenoid
[tex]U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3[/tex]
