Answer:
104
Explanation:
U = Energy stored in the solenoid = 6.00 μJ = 6.00 x 10⁻⁶ J
i = current flowing through the solenoid = 0.4 A
L = inductance of the solenoid
Energy stored in the solenoid is given as
U = (0.5) L i²
6.00 x 10⁻⁶ = (0.5) L (0.4)²
L = 75 x 10⁻⁶
Inductance is given as
l = length of the solenoid = 0.7 m
N = number of turns
r = radius = 5.00 cm = 0.05 m
Area of cross-section is given as
A = πr²
A = (3.14) (0.05)²
A = 0.00785 m²
Inductance is given as
[tex]L=\frac{\mu _{o}N^{2}A}{l}[/tex]
[tex]75\times 10^{-6}=\frac{(12.56\times 10^{-7})N^{2}(0.00785)}{0.7}[/tex]
N = 73
Winding density is given as
density = n = [tex]\frac{N}{l}[/tex]
n = [tex]\frac{73}{0.7}[/tex]
n = 104
