The energy stored in the magnetic field of a solenoid 30

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The energy stored in the magnetic field of a solenoid 30 cm long and 3 cm diameter wound with 1000 turns of wire carrying a current of 10 A is—

1. 0.015 joule
2. 0.15 joule
3. 0.5 joule
4. 1.15 joule

Correct Option: B

Since we know that,
L = N² × ρ
given
N = 1000 turns diameter
= 3 cm, L = 30 cm.

ρ =	µ₀A	=	4π × 10^–7 × π × (1.5 × 10^–2)²
	L		0.3

= 29.6088 × 10^–10
L = 29.6088 × 10^–10 × (10³)²
= 29.6088 × 10^–4 H
where,
L = length of solenoid
W = 1/2 LI²
= 1/2 × 29.6088 × 10^–4 × (10)²
= 0.148 ≈ 0.15 J.