NEET-XII-Physics
38: Electromagnetic Induction
- #12radiuA 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.Ans : 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
`` q=I∆t=\frac{4{\,\mathrm{\,\pi \,}}^{2}\times {10}^{-4}}{20\times dt}\times ∆t``
`` =2\times {10}^{-4}\,\mathrm{\,C\,}``
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