Answer:
B = 32.17 x 10^-8 Tesla
u = 8.24 x 10^-8 J/m^3
P/A = 24.72 W/m^2
Explanation:
E = 96.5 V/m
velocity of light, c = 3 x 10^8 m/s
Let B be the magnetic field.
The relation between the electric field strength and the magnetic field strength is given by
B = E / c = 96.5 / (3 x 10^8) = 32.17 x 10^-8 Tesla
Let u be the energy density.
u = 8.24 x 10^-8 J/m^3
Let Power flow per unit area is
P/A = u x c = 8.24 x 10^-8 x 3 x 10^8 = 24.72 W/m^2
Answer:
The air fraction to be removed is 0.11
Given:
Initial temperature, T = = 283 K
Pressure, P = 250 kPa
Finally its temperature increases, T' = = 318 K
Solution:
Using the ideal gas equation:
PV = mRT
where
P = Pressure
V = Volume
m = no. of moles of gas
R = Rydberg's Constant
T = Temperature
Now,
Considering the eqn at constant volume and pressure, we get:
mT = m'T'
Thus
(1)
Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:
From eqn (1):