Answer: 13.1 μH
Explanation:
Given
length of heating coil, l = 1 m
Diameter of heating coil, d = 0.8 cm = 8*10^-3 m
No of loops, N = 400
L = μN²A / l
where
μ = 4π*10^-7 = 1.26*10^-6 T
A = πd²/4 = (π * .008 * .008) / 4 = 6.4*10^-5 m²
L = μN²A / l
L = [1.26*10^-6 * 400 * 400* 6.5*10^-5] / 1
L = 1.26*10^-6 * 1.6*10^5 * 6.5*10^-5
L = 1.31*10^-5
L = 13.1 μH
Thus, from the calculations above, we can say that the total self inductance of the solenoid is 13.1 μH
Answer:
Their total self-inductance assuming they act like a single solenoid is 10.11 μH
Explanation:
Given;
diameter of the heating coil, d = 0.800 cm
combined length of heating coil and hair dryer, [tex]l[/tex] = 1.0 m
number of turns, N = 400 turns
Formula for self-inductance is given as;
[tex]L = \frac{\mu_oN^2A}{l}[/tex]
where
μ₀ is constant = 4π x 10⁻⁷ T.m/A
A is the area of the coil:
A = πd²/4
A = π (0.8 x 10⁻²)²/4
A = 5.027 x 10⁻⁵ m²
[tex]L = \frac{\mu_oN^2 A}{l } = \frac{4\pi *10^{-7}(400)^2 *5.027*10^{-5}}{1 } \\\\L =1.011 *10^{-5} \ H\\\\L = 10.11 \mu H[/tex]
Therefore, their total self-inductance assuming they act like a single solenoid is 10.11 μH
