Question

radiuA long solenoid ofs 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,
$e=\frac{∆\mathrm{\varphi }}{∆t}=\frac{4{\mathrm{\pi }}^2\times {10}^{-4}}{∆t}$
The current through the solenoid is given by
$I=\frac{e}{R}=\frac{4{\mathrm{\pi }}^2\times {10}^{-4}}{∆t\times 20}$
The charge flown through the galvanometer is given by
$$\begin{gathered} q=I∆t=\frac{4{\mathrm{\pi }}^2\times {10}^{-4}}{20\times dt}\times ∆t \\ =2\times {10}^{-4}\mathrm{C} \end{gathered}$$