An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
arge density σ = -0.35 μC/m2. A thin wire, with linear charge density λ = 1.1 μC/m, is inserted along the shells' axis. The shell and the wire do not touch and these is no charge exchanged between them.a) What is the new surface charge density, in microcoulombs per square meter, on the inner surface of the cylindrical shell?b) What is the new surface charge density, in microcoulombs per square meter, on the outer surface of the cylindrical shell?c) Enter an expression for the magnitude of the electric field ou
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Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0
This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u /><u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u /><u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
Answer:
(a) Magnetic moment will be
(b) Torque will be
Explanation:
We have given dimension of the rectangular 5.4 cm × 8.5 cm
So area of the rectangular coil
Current is given as
Number of turns N = 25
(A) We know that magnetic moment is given by
(b) Magnetic field is given as B = 0.350 T
We know that torque is given by