A 1800-turn solenoid has a cross-sectional area equal to 2.0 cm² and a length equal to 30 cm. The solenoid carries a current of 4.0 A. (a) Calculate the magnetic energy stored in the solenoid using $U = frac{1}{2}LI^2$, where $L = mu_0n^2Al$. mJ (b) Divide your answer in part (a) by the volume of the region inside the solenoid to find the magnetic energy per unit volume in the solenoid. kJ/m³ (c) Check your Part (b) result by computing the magnetic energy density from $u_m = frac{B^2}{2mu_0}$ where $B = mu_0nI$. kJ/m³
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0 cm^2 = \(4.0 \times 10^{-4}\) m^2 Length, L = 30 cm = 0.3 m Current, I = 4.0 A Substitute the values into the formula: \(U = \frac{1}{2} \times (4\pi \times 10^{-7} \times 1800 \times 4.0 \times 10^{-4} \times 4.0^2) \times 0.3\) Show more…
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