Question

A solenoid 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)

Answer 1

Answer:

N/l = 104

Explanation:

Energy stored in the inductor is given by the formula

$U = \frac{1}{2}Li^2$

now we have

$6\times 10^{-6} = \frac{1}{2}L(0.400)^2$

now we have

$L = 7.5 \times 10^{-5}$

now we have

$L = \frac{\mu_0 N^2 \pi r^2}{l}$

$7.5 \times 10^{-5} = \frac{4\pi \times 10^{-7} N^2 \pi(0.05)^2}{0.7}$

$N = 73 turns$

now winding density is turns per unit length

$N/l = 104$