Question

A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm having 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.

Answer 1

Given:-

Radius of the solenoid, r = 2 cm = 2 × 10⁻² m

Number of turns per centimetre, n = 100 = 10000 turns/m

Current flowing through the coil, i = 5 A

The magnetic field through the solenoid is given by

B = μ₀ni = 4π × 10⁻⁷ × 10000 × 5

= 20π × 10⁻³ T

Flux linking with per turn of the second solenoid = Bπr² = Bπ × 10⁻⁴

Total flux linking the second coil, ϕ₁ = Bn₂πr²

∴ ϕ₁ = 100 × π × 10⁻⁴ × 20π × 10⁻³

When the direction of the current is reversed, the total flux linking the second coil is given by

ϕ₂ = −Bn₂πr²

= −(100 × π × 10⁻⁴ × 20π × 10⁻³ )

The change in the flux through the second coil is given by

Δϕ = ϕ₂ − ϕ₁

= 2 × (100 × π × 10⁻⁴ × 20π × 10⁻³)

Now,

\[e = \frac{∆ \phi}{∆ t} = \frac{4 \pi^2 \times {10}^{- 4}}{∆ t}\]

The current through the solenoid is given by

\[I = \frac{e}{R} = \frac{4 \pi^2 \times {10}^{- 4}}{∆ t \times 20}\]

The charge flown through the galvanometer is given by

\[q = I ∆ t = \frac{4 \pi^2 \times {10}^{- 4}}{20 \times dt} \times  ∆ t\]

\[ = 2 \times  {10}^{- 4}   C\]