An air core solenoid with 68 turns is 8 cm long and has a diameter of 1.2 cm. how much energy is stored in its magnetic field when it carries a current of 0.77a?

Question

Izabelle Fuentes

Physics

29 September, 07:24

An air core solenoid with 68 turns is 8 cm long and has a diameter of 1.2 cm. how much energy is stored in its magnetic field when it carries a current of 0.77a?

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Ismael Delgado · Answer 1

To determine the energy stored in the magnetic field, use the formula: E = ½ LI2 where E is the energy stored, L is the inductance (units: Henry), and I is the current (units: Ampere).

Correspondingly, L = (µN2A) / l, where, µ = permeability of medium (in this case, air), N = no. of turns, A = cross-sectional area, and l = length. Given: N = 68 turns, l = 8 cm, d = 1.2 cm, and I = 0.77 A, other necessary variables are A = 1.13 cm2 and µair = 1.2566 x 10-6 H/m. Applying the formula and checking that the units agree with each other, the final answer is 2.4352 x 10-6 J.