A solenoid of 1000 turns is wound with wire of diameter 0.1 cm and has

1.	A solenoid of 1000 turns is wound with wire of diameter 0.1 cm and has a self inductance of 2.4 π × 10-5 H. Find (a) the cross-sectional area of the solenoid (b) the magnetic flux through one turn of the solenoid when a current of 3 A flows through it.
Answer» Data: N = 1000, D = 0.1 cm, L = 2.4π × 10^-5H I = 3A, μ₀= 4π × 10^-7 H/m The number of turns per unit length. n = \(\cfrac1{1\,mm}\) = 1 mm^-1 = 10³ m^-1 and the length of the solenoid, l = ND = 1000 × 0.1 = 100 cm = 1 m L = μ₀n² lA (a) The area of cross section A = \(\cfrac{L}{μ_ 0n^2l}\) = \(\cfrac{2.4π×10^{-5}}{(4π×10^{-7} )(10^3)^2 (1)}\) = \(\cfrac{24π}{4π}\) x 10^-5 = 6 × 10^-5 m² (b) Magnetic flux through one turn, Φ_m = BA = (μ₀nI)A = (4π × 10^-7)(10³)(3)(6 × 10^-5) = 72π × 10^-9 Wb

A solenoid of 1000 turns is wound with wire of diameter 0.1 cm and has a self inductance of 2.4 π × 10-5 H. Find (a) the cross-sectional area of the solenoid (b) the magnetic flux through one turn of the solenoid when a current of 3 A flows through it.

Answer»

Data: N = 1000, D = 0.1 cm, L = 2.4π × 10^-5H

I = 3A, μ₀= 4π × 10^-7 H/m

The number of turns per unit length.

n = \(\cfrac1{1\,mm}\) = 1 mm^-1 = 10³ m^-1

and the length of the solenoid,

l = ND = 1000 × 0.1 = 100 cm = 1 m

L = μ₀n² lA

(a) The area of cross section

A = \(\cfrac{L}{μ_ 0n^2l}\) = \(\cfrac{2.4π×10^{-5}}{(4π×10^{-7} )(10^3)^2 (1)}\) = \(\cfrac{24π}{4π}\) x 10^-5

= 6 × 10^-5 m²

(b) Magnetic flux through one turn,

Φ_m = BA = (μ₀nI)A

= (4π × 10^-7)(10³)(3)(6 × 10^-5)

= 72π × 10^-9 Wb

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