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
The freezing point ..... :)
According to periodic trends in the periodic table, the atomic radius decreases from left to right.
In period three, the element with the smallest atomic radius would be the element in the rightmost area. Protons increase as it goes to the right, which would mean they pull in electrons closer which decreases the size.
So in period 3, the element with the smallest atomic radius is Argon (Ar).
