Respuesta :
Explanation:
(a) We have,
Length of solenoid, l = 55 cm = 0.55 m
Diameter of the solenoid, d = 10 cm
Radius, r = 5 cm = 0.05 m
Number of loops in the solenoid is 1000.
(a) The self inductance in the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2A}{l}[/tex]
A is area
[tex]L=\dfrac{4\pi \times 10^{-7}\times (1000)^2\times \pi (0.05)^2}{0.55}\\\\L=0.0179\ H\\\\L=17.9\ mH[/tex]
(b) The energy stored in the inductor is given by :
[tex]E=\dfrac{1}{2}LI^2\\\\E=\dfrac{1}{2}\times 0.0179\times (19.5)^2\\\\E=3.4\ J[/tex]
Hence, this is the required solution.
a. The self-inductance (in mH) should be 17.9 mH.
b. The energy should be stored should be 3.4 J.
Calculation of the self-inductance and energy:
Since
Length of solenoid, l = 55 cm = 0.55 m
Diameter of the solenoid, d = 10 cm
Radius, r = 5 cm = 0.05 m
Number of loops in the solenoid is 1000.
a.
Now the self inductance should be
= 4π *10^*7*(1000)^2 * π(0.05)^2/ 0.55
= 0.0179 H
= 17.9 mH
b.
Now the energy should be
= 1/2 * 0.0179 * (19.5)^2
= 3.4 J
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