Answer:
E = 0.1472 J
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 :
[tex]E=\dfrac{1}{2}LI^2[/tex]
Where
L is self inductance of the solenoid,
[tex]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[/tex]
So,
[tex]E=\dfrac{1}{2}\times 0.0184\times 4^2\\\\E=0.1472\ J[/tex]
So, 0.1472 J of energy is stored in the solenoid.