Answer:
5.92 x 10⁻⁵ J
Explanation:
[tex]N[/tex] = Total number of turns
[tex]d[/tex] = diameter of insulated wire = 1 mm = 0.001 m
[tex]l[/tex] = length of the cylindrical core = 30 cm = 0.30 m
Length is given as
[tex]l[/tex] = [tex]N[/tex] [tex]d[/tex]
0.30 = [tex]N[/tex] (0.001)
[tex]N[/tex] = 300
[tex]r[/tex] = radius of the core = 5 cm = 0.05 m
Area is given as
[tex]A[/tex] = πr²
[tex]A[/tex] = (3.14) (0.05)²
[tex]A[/tex] = 0.00785 m²
Inductance of the solenoid is given as
[tex]L = \frac{\mu _{o}N^{2}A}{l}[/tex]
[tex]L = \frac{(12.56\times 10^{-7}))(300)^{2}(0.00785)}{0.30}[/tex]
[tex]L [/tex] = 0.00296
[tex]i [/tex] = current in the solenoid = 0.20 A
Energy stored in the solenoid is given as
[tex]U = (0.5)Li^{2}[/tex]
[tex]U = (0.5)(0.00296)(0.20)^{2}[/tex]
[tex]U [/tex] = 5.92 x 10⁻⁵ J
