Answer:
Explanation:
We shall solve this question with the help of Ampere's circuital law.
Ampere's ,law
∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire
we shall find magnetic field at distance x . current enclosed in the area of circle of radius x
= I x π x² / π R²
= I x² / R²
B x 2π x = μ₀ x current enclosed
B x 2π x = μ₀ x I x² / R²
B = μ₀ I x / 2π R²
Maximum magnetic B₀ field will be when x = R
B₀ = μ₀I / 2π R
Given
B = B₀ / 3
μ₀ I x / 2π R² = μ₀I / 2π R x 3
x = R / 3
b ) The largest value of magnetic field is on the surface of wire
B₀ = μ₀I / 2π R
At distance x outside , let magnetic field be B
Applying Ampere's circuital law
∫ B dl = μ₀ I
B x 2π x = μ₀ I
B = μ₀ I / 2π x
Given B = B₀ / 3
μ₀ I / 2π x = μ₀I / 2π R x 3
x = 3R .
Energy in a spring:
E = 0.5 * k * x²
k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m
E = 0.5 * 800 * 0.05² = 1
Answer:
hope this answer helps you.
Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called
cavitation.
So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at

is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.