Explanation:
It is given that,
Length of solenoid, l = 28 cm = 0.28 m
Area of cross section, A = 0.475 cm² = 4.75 × 10⁻⁵ m²
Current, I = 85 A
(a) The inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_oN^2A}{l}[/tex]
[tex]L=\dfrac{4\pi\times 10^{-7}\times (645)^2\times 4.75\times 10^{-5}}{0.28}[/tex]
L = 0.0000886 H
[tex]L=8.86\times 10^{-5}\ H[/tex]
(b) Energy contained in the coil's magnetic field is given by :
[tex]U=\dfrac{1}{2}LI^2[/tex]
[tex]U=\dfrac{1}{2}\times 8.86\times 10^{-5}\ H\times (85\ A)^2[/tex]
U = 0.3200675 Joules
or
U = 0.321 Joules
Hence, this is the required solution.
