Answer:
a. B = 0.20T
b. u = 17230.6 J/m³
c. E = 0.236J
d. L = 5.84*10^-5 H
Explanation:
a. In order to calculate the magnetic field in the solenoid you use the following formula:
[tex]B=\frac{\mu_o n i}{L}[/tex] (1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
n: turns of the solenoid = 460
L: length of the solenoid = 25.0cm = 0.25m
i: current = 90.0A
You replace the values of the parameters in the equation (1):
[tex]B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T[/tex]
The magnetic field in the solenoid is 0.20T
b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:
[tex]u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}[/tex]
The energy density is 17230.6 J/m³
c. The total energy contained in the solenoid is:
[tex]E=uV[/tex] (2)
V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:
[tex]V=AL[/tex]
A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2
[tex]V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3[/tex]
Then, the energy contained in the solenoid is:
[tex]E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J[/tex]
The energy contained is 0.236J
d. The inductance of the solenoid is calculated as follow:
[tex]L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H[/tex]
The inductance of the solenoid is 5.84*10^-5 H
