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: If the force stays the same, the acceleration would decrease
Answer:
A. 70 m OS the correct one
The convection going on with the magma in the asthenosphere
