A 0.20-cm radius cylinder, 3.0 cm long, is wrapped with wire to form an inductor. At the instant the magnetic field in the interior is 5.0 mT the energy stored in the field is:

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The energy stored in a magnetic field is given by:$$U = \frac{1}{2}LI^2$$where L is the inductance of the inductor and I is the current flowing through it. The inductance of a solenoid is given by:$$L = \frac{\mu_0n^2A}{l}$$where \mu_0 is the permeability of free space, n is the number of turns per unit length, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.In this problem, we have:$$r = 0.20 	ext{ cm} = 2.0 	imes 10^{-3} 	ext{ m}$$$$l = 3.0 	ext{ cm} = 3.0 	imes 10^{-2} 	ext{ m}$$$$B = 5.0 	ext{ mT} = 5.0 	imes 10^{-3} 	ext{ T}$$$$\mu_0 = 4\pi 	imes 10^{-7} 	ext{ T$\cdot$m/A}$$The number of turns per unit length is:$$n = \frac{N}{l}$$where N is the total number of turns. We don't know the value of N, but we can use the fact that the magnetic field inside a solenoid is given by:$$B = \mu_0nI$$to solve for n:$$n = \frac{B}{\mu_0I}$$Substituting this into the equation for the inductance, we get:$$L = \frac{\mu_0B^2l}{\mu_0^2I^2}$$Simplifying, we get:$$L = \frac{B^2l}{\mu_0I^2}$$Substituting the given values, we get:$$L = \frac{(5.0 	imes 10^{-3} 	ext{ T})^2(3.0 	imes 10^{-2} 	ext{ m})}{4\pi 	imes 10^{-7} 	ext{ T$\cdot$m/A}}$$$$L = 3.8 	imes 10^{-6} 	ext{ H}$$The energy stored in the magnetic field is:$$U = \frac{1}{2}LI^2 = \frac{1}{2}(3.8 	imes 10^{-6} 	ext{ H})I^2$$We don't know the value of I, but we can use the fact that the magnetic field inside a solenoid is given by:$$B = \mu_0nI$$to solve for I:$$I = \frac{B}{\mu_0n}$$Substituting this into the equation for the energy, we get:$$U = \frac{1}{2}\left(\frac{B^2l}{\mu_0n^2}ight)\left(\frac{B}{\mu_0n}ight)^2$$Simplifying, we get:$$U = \frac{B^4l}{2\mu_0^3n^3}$$Substituting the given values, we get:$$U = \frac{(5.0 	imes 10^{-3} 	ext{ T})^4(3.0 	imes 10^{-2} 	ext{ m})}{2(4\pi 	imes 10^{-7} 	ext{ T$\cdot$m/A})^3}$$$$U = 3.8 	imes 10^{-6} 	ext{ J}$$Therefore, the energy stored in the magnetic field is 3.8 	imes 10^{-6} J.

A 020-Cm Radius Cylinder, 3