# A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume that the loop is perpendicular to the magnetic field. Part A What is the average induced emf?

1 answer

###### Question:

A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume that the loop is perpendicular to the magnetic field. Part A What is the average induced emf?

## Answers

Answer:0.021 VExplanation:The average induced emf (E) can be calculated usgin the Faraday's Law:[tex] E = - \frac{N*\Delta \phi}{\Delta t} [/tex] Where:N = is the number of turns = 1 ΔΦ = ΔB*A Δt = is the time = 0.3 s A = is the loop of wire area = πr² = πd²/4 ΔB: is the magnetic field = (0 - 1.04) T Hence the average induced emf is:[tex] E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V [/tex] Therefore, the average induced emf is 0.021 V.I hope it helps you!

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