The magnetic field perpendicular to a single 13.2-cm diameter circular loop of copper wire decreases uniformly from 0.670 T to zero. If the

Question

The magnetic field perpendicular to a single 13.2-cm diameter circular loop of copper wire decreases uniformly from 0.670 T to zero. If the wire is 2.25 mm in diameter, how much charge moves past a point in the coil during this operation? (rhoCu = 1.68 x 10-8 Ω.m)

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Amity 3 years 2021-08-12T21:29:09+00:00 1 Answers 13 views 0

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  1. mocmien2
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    2021-08-12T21:31:00+00:00
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    Answer:

    5.23 C

    Explanation:

    The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.

    Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ – B₁

    B₁ = 0.670 T and B₂ = 0 T

    ΔB = B₂ – B₁ = 0 – 0.670 T = – 0.670 T

    A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m

    A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²

    ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt

    Now, the resistance R of the circular wire R = ρl/A’ where ρ = resistivity of copper wire =  1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A’ = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m

    R = ρl/A’ = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω

    So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω  

    IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C

    Since ΔQ = It = 5.232 C ≅ 5.23 C

    So the charge is 5.23 C

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