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: A)0 J B)3.8*10^-6 J C)7.5 *10^-6 J D)7.5*10^-4J E)9.9 J

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The energy stored in a magnetic field is given by:$$U = \frac{1}{2}LI^2$$where:* U is the energy stored in joules (J)* L is the inductance in henrys (H)* I is the current in amperes (A)The inductance of a solenoid is given by:$$L = \frac{\mu_0N^2A}{l}$$where:* L is the inductance in henrys (H)* $\mu_0$ is the permeability of free space (4π × 10^-7 H/m)* N is the number of turns* A is the cross-sectional area of the solenoid in square meters (m^2)* l is the length of the solenoid in meters (m)In this problem, we are given the following information:* r = 0.20 cm = 2.0 × 10^-3 m* l = 3.0 cm = 3.0 × 10^-2 m* B = 5.0 mT = 5.0 × 10^-3 TWe can calculate the cross-sectional area of the solenoid as follows:$$A = \pi r^2 = \pi (2.0 × 10^-3 	ext{ m})^2 = 1.26 × 10^-5 	ext{ m}^2$$We can calculate the inductance of the solenoid as follows:$$L = \frac{\mu_0N^2A}{l} = \frac{(4\pi 	imes 10^{-7} 	ext{ H/m})(N^2)(1.26 	imes 10^{-5} 	ext{ m}^2)}{3.0 	imes 10^{-2} 	ext{ m}}$$We do not know the number of turns, N, so we cannot calculate the inductance exactly. However, we can use the given information to set up an equation for the inductance:$$L = \frac{(4\pi 	imes 10^{-7} 	ext{ H/m})(N^2)(1.26 	imes 10^{-5} 	ext{ m}^2)}{3.0 	imes 10^{-2} 	ext{ m}}$$$$L = 5.31 	imes 10^{-11} 	ext{ H}N^2$$We can now calculate the energy stored in the magnetic field as follows:$$U = \frac{1}{2}LI^2 = \frac{1}{2}(5.31 	imes 10^{-11} 	ext{ H}N^2)I^2$$We do not know the current, I, so we cannot calculate the energy stored exactly. However, we can use the given information to set up an equation for the energy stored:$$U = \frac{1}{2}(5.31 	imes 10^{-11} 	ext{ H}N^2)I^2$$$$U = 2.66 	imes 10^{-11} 	ext{ J}N^2I^2$$We can now use the given information to eliminate the unknown variables and solve for the energy stored.$$B = \frac{\mu_0NI}{l}$$$$I = \frac{Bl}{\mu_0N}$$$$U = 2.66 	imes 10^{-11} 	ext{ J}N^2\left(\frac{Bl}{\mu_0N}ight)^2$$$$U = 2.66 	imes 10^{-11} 	ext{ J}\frac{B^2l^2}{\mu_0^2}$$$$U = 2.66 	imes 10^{-11} 	ext{ J}\frac{(5.0 	imes 10^{-3} 	ext{ T})^2(3.0 	imes 10^{-2} 	ext{ m})^2}{(4\pi 	imes 10^{-7} 	ext{ H/m})^2}$$$$U = 3.8 	imes 10^{-6} 	ext{ J}$$Therefore, the energy stored in the magnetic field is 3.8 × 10^-6 J.

A 020-Cm Radius Cylinder, 3