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A solenoid of 2100 turns, area 10 cm2, and length 30 cm carries a current of 4.0 A. (a) Calculate the magnetic energy stored in

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Explanation:

Given that,

The number of turns in the solenoid, N = 2100

Area of the solenoid, A = 10 cm² = 0.001 m²

The length of the solenoid, l = 30 cm = 0.3 m

Current in the solenoid, I = 4 A

We need to find the magnetic energy stored in the solenoid. The expression for the stored energy is :

$E=\dfrac{1}{2}LI^2$

Where

L is self inductance of the solenoid,

$L=\dfrac{\mu_oN^2A}{l}\\\\L=\dfrac{4\pi \times 10^{-7}\times 2100^2\times 0.001}{0.3}\\\\L=0.0184\ H$

So,

$E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J$

So, 0.1472 J of energy is stored in the solenoid.

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