Question

20. How much charge will flow through a 2002 galvanometer

Answer 1

Answer:

5.76 μC

Explanation:

The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T

So, ε = -ΔΦ/Δt

ε = -NAΔB/Δt

ε = -NAΔB/Δt

Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)

So, iR = -NAΔB/Δt

iΔt = -NAΔB/R

Δq = -NAΔB/R where Δq = charge = iΔt

substituting the values of the variables into the equation, we have

Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω

Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω

Δq = 0.011π × 10⁻¹ m²T/600 Ω

Δq = 0.03456 × 10⁻¹ m²T/600 Ω

Δq = 5.76 × 10⁻⁶ C

Δq = 5.76 μC