Explanation:
It is given that,
Length of the tube, l = 33 cm = 0.33 m
Diameter of the tube, d = 8.1 cm = 0.081 m
Number of turns, N = 590
(a) We need to find the self- inductance of the solenoid. The formula for the self inductance is given by :
[tex]L=\dfrac{\mu_o N^2A}{l}[/tex]
A is the area of the solenoid
[tex]L=\dfrac{4\pi \times 10^{-7}\times (590)^2\times \pi \times (0.0405)^2}{0.33}[/tex]
L = 0.00683 H
or
L = 6.83 mH
So, the self inductance of the solenoid is 6.83 mH.
(b) The current in this solenoid increases at the rate of 3 A/s. [tex]\dfrac{dI}{dt}=3\ A/s[/tex]
EMF in the solenoid is given by :
[tex]E=-L\dfrac{dI}{dt}[/tex]
[tex]E=-6.83\times 3[/tex]
E = -20.49 volt
Hence, this is the required solution.
