Question

A solenoid of length 20 cm, area of cross-section 4.0 cm² and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-sectional area 8.0 cm² and  length 10 cm. Find the mutual inductance between the solenoids.

Answer 1

Mutual inductance
M = μ₀N₁N₂πr₁²l
= 4π × 10⁻⁷ × 4 × 10³ × 2 × 10³ × 4 × 10⁻⁴ × 10 × 10⁻²
= 0.04 × 10⁻² H

Given that,
For solenoid-1
Area of cross section, a₁ = 4 cm² = 4 × 10⁻⁴ m²
Length of the solenoid, l₁ = 20 cm = 0.20 m
Number of turns per unit length , n₁ = 4000/0.2 m = 20000 turns/m

For solenoid-2
Area of cross section, a₂ = 8 cm² = 4 × 10⁻⁴ m²
Length of the solenoid, l₂ = 10 cm = 0.1 m
Number of turns per unit length, n₂ = 2000/0.1 m = 20000 turns/m

It is given that the solenoid-1 is placed inside the solenoid-2

Let the current through the solenoid-2 be i.

The magnetic field due to current in solenoid-2

B = μ₀n₂i = \[\left( 4\pi \times {10}^{- 7} \right) \times \left( 20000 \right) \times i\]

Now,

Flux through the coil-1 is given by

ϕ = n₁l₁.B.a₁ = n₁l₁(μ₀n₂i) × a₁

\[\phi = 2000 \times 20000 \times 4\pi \times  {10}^{- 7}  \times i \times 4 \times  {10}^{- 4}\]

If the current flowing in the coil-2 changes, then emf is induced in the coil-1

Thus, emf induced in the coil-1 due to change in the current in coil-2 is given by

`E=(dphi)/(dt)=64pixx10^-4xx(di)/(dt)`

Now, `M=E/"di/dt"=64pixx10^-4H=2xx10^-2H .......["As E = Mdi/dt"]`