Answer:
M' = μ₀n₁n₂πr₂²
Explanation:
Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.
So, M = N₂Ф₂₁/i₁
substituting the values of the variables into the equation, we have
M = N₂Ф₂₁/i₁
M = N₂B₁A₂/i₁
M = n₂lμ₀n₁i₁πr₂²/i₁
M = lμ₀n₁n₂πr₂²
So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²
M' = μ₀n₁n₂πr₂²
Answer:
199.0521 Will be the answer
Answer:
8.27°
Explanation:
To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.
![d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m](https://tex.z-dn.net/?f=d%3Dy_1-y_2%3D%5Cfrac%7BF_1%7D%7Bk_1%7D-%5Cfrac%7BF_2%7D%7Bk_2%7D%3D%5Cfrac%7B0.5mg%7D%7B31N%2Fm%7D-%5Cfrac%7B0.5mg%7D%7B63N%2Fm%7D%5C%5C%5C%5Cd%3D0.5%281.6kg%29%289.8m%2Fs%5E2%29%5B%5Cfrac%7B1%7D%7B31N%2Fm%7D-%5Cfrac%7B1%7D%7B63N%2Fm%7D%5D%3D0.128m)
by a simple trigonometric relation you obtain that the angle:
hence, the angle between the rod and the horizontal is 8.27°
