An iron ring 30 cm mean diameter is made of square of iron of 2 cm ×

1.	An iron ring 30 cm mean diameter is made of square of iron of 2 cm × 2 cm cross section and is uniformly wound with 400 turns of wire of 2 mm2 cross-section. Calculate the value of the self-inductance of the coil. Assume μr = 800.
Answer» L = μ₀ μ_r AN²/l. Here N = 400 ; A = 2 × 2 = 4 cm² = 4 × 10⁻⁴ m² ; l = 0.3π m ; μ_r = 800 ∴ L = 4π × 10⁻⁷ × 800 × 4 × 10⁻⁴ (400)²/0.3π = 68.3 mH Note. The cross-section of the wire is not relevant to the given question. Third Method for L It will be (i) above that L = NΦ/I ∴ N Φ = LI or −NΦ = −L I Differentiating both sides, we get - d/dt(NΦ) = - L x dI/dt (assuming L to be constant) ; - N x dΦ/dt = - L x dI/dt - N x dΦ/dt = self-induced e.m.f. ∴ e_L= - L x dI/dt If dI/dt = 1 ampere/second and e_L = 1 volt, then L = 1 H

An iron ring 30 cm mean diameter is made of square of iron of 2 cm × 2 cm cross section and is uniformly wound with 400 turns of wire of 2 mm2 cross-section. Calculate the value of the self-inductance of the coil. Assume μr = 800.

Answer»

L = μ₀ μ_r AN²/l.

Here N = 400 ; A = 2 × 2 = 4 cm² = 4 × 10⁻⁴ m² ; l = 0.3π m ; μ_r = 800

∴ L = 4π × 10⁻⁷ × 800 × 4 × 10⁻⁴ (400)²/0.3π = 68.3 mH

Note. The cross-section of the wire is not relevant to the given question.

Third Method for L

It will be (i) above that L = NΦ/I

∴ N Φ = LI or −NΦ = −L I

Differentiating both sides, we get - d/dt(NΦ) = - L x dI/dt (assuming L to be constant) ;

- N x dΦ/dt = - L x dI/dt

- N x dΦ/dt = self-induced e.m.f.

∴ e_L= - L x dI/dt

If dI/dt = 1 ampere/second and e_L = 1 volt, then L = 1 H

Discussion

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