Asolenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µj of energy when a 0.400-a current runs through it. what is the winding density of the solenoid? (μ 0 = 4π × 10-7 t · m/a)

N/l = 104

Explanation:

Energy stored in the inductor is given by the formula

now we have

now we have

now we have

now winding density is turns per unit length

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

N = 73

Winding density is given as

density = n =

n =

n = 104