Cylindrical magnets

Calculate the magnetic field energy of a cylindrical coil with 400 turns, a length of 0.4 m, and a radius of 20 mm. A current of 3A passes through the coil. (µo = 4π 10^-7 H/m)

Correct answer:

E =  0.0007 J

Step-by-step explanation:

$$\begin{gathered} N=400 \\ l=0.4\text{m} \\ D=20\text{mm}\to \text{m}=20:1000\text{m}=0.02\text{m} \\ I=3\text{A} \\ \mu =4\pi \cdot 10^{-7}=4\cdot 3.1416\cdot 10^{-7}\doteq 1.2566\cdot 10^{-6}\text{H/m} \\ r=D/2=0.02/2=\frac{1}{100}=0.01\text{m} \\ S=\pi \cdot r^2=3.1416\cdot 0.01^2\doteq 0.0003{\text{m}}^2 \\ L=N^2\cdot \mu \cdot S/l=400^2\cdot 1.2566\cdot 10^{-6}\cdot 0.0003/0.4\doteq 0.0002\text{H} \\ E=\frac{1}{2}\cdot L\cdot I^2=\frac{1}{2}\cdot 0.0002\cdot 3^2=0.0007\text{J} \end{gathered}$$

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