Answer:
A. 3.9 V B. 1.9 fA
Step-by-step explanation:
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time
ε = ΔΦ/Δt
ε = ΔABcos0°/Δt
ε = AΔB/Δt
A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T. ΔB = B₁ - B₀ = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.
So, ε = AΔB/Δt
ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s
ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s
ε = 3.926 V
ε ≅ 3.9 V
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².
So, i = ε/R
= ε/ρl/A
= εA/ρl
= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)
= 15.704 × 10⁻⁶ V-m²/(82.9356 × 10⁸ Ωm²
= 0.1894 × 10⁻¹⁴ A
= 1.894 × 10⁻¹⁵ A
≅ 1.9 fA