Complete Question:
In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.
a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.
b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.
c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.
d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?
Answer:
force, 1.318 ₓ 10⁻⁴
direction, 18.435°
Explanation:
The attached file gives a breakdown step by step solution to the questions
Answer:
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Explanation:
Answer:
∆T = Mv^2Y/2Cp
Explanation:
Formula for Kinetic energy of the vessel = 1/2mv^2
Increase in internal energy Δu = nCVΔT
where n is the number of moles of the gas in vessel.
When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas
We say
1/2mv^2 = ∆u
1/2mv^2 = nCv∆T
Since n = m/M
1/2mv^2 = mCv∆T/M
Making ∆T subject of the formula we have
∆T = Mv^2/2Cv
Multiple the RHS by Cp/Cp
∆T = Mv^2/2Cv *Cp/Cp
Since Y = Cp/CV
∆T = Mv^2Y/2Cp k
Since CV = R/Y - 1
We could also have
∆T = Mv^2(Y - 1)/2R k
Answer:
The electron cloud is mostly empty space