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`         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, find the charge flown through the galvanometer.`
5 years ago

```							Sol. r = 2 cm = 2 x 10^–2 m
n = 100 turns / cm = 10000 turns/m
i = 5 A
B = μ0 ni
= 4π x 10^–7  x 10000 x 5 = 20π x 10^–3 = 62.8 x 10^–3 T
n base 2 = 100 turns
R = 20 Ω
r = 1 cm = 10^–2 m
Flux linking per turn of the second coil = Bπ r^2 = Bπ x 10^–4
ϕ base 1 = Total flux linking = Bn base 2 π r2 = 100 x π x 10^–4 x 20π x 10^–3
When current is reversed.
ϕ base 2 = – ϕ  base 1
dϕ = ϕ base 2 – ϕ base 1 = 2 x 100 x π x 10^–4 x 20π x 10–3
E = dϕ/dt = 4π^2 x 10^-4/dt
l = E/R = 4π^2 x 10^-4/dt x 20
q = idt = 4π^2 x 10^-4/dt x 20 x dt = 2 x 10^-4 C.

```
5 years ago
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