A flat square coil of wire with 15 turns and an area of 0.40 m2 is placed with the plane of its area parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in a time of 0.050 s. What is the magnitude of the average induced emf in the coil?

Question

A flat square coil of wire with 15 turns and an area of 0.40 m2 is placed with the plane of its area parallel to a magnetic field of 0.75 T.  The coil is flipped so its plane is perpendicular to the magnetic field in a time of 0.050 s.  What is the magnitude of the average induced emf in the coil?

BlackTom AI · Accepted Answer

Let’s restate the problem clearly: A flat square coil has N = 15 turns and area A = 0.40 m^2. It sits with its plane parallel to a magnetic field B = 0.75 T. The coil is flipped so that its plane becomes perpendicular to the field in Δt = 0.050 s. We’re asked for the magnitude of the average induced emf.

First, evaluate the magnetic flux through the coil before and after flipping. The magnetic flux is Φ = B · A · cos(θ), where θ is the angle between the field and the coil’s normal.
- Initially, when the plane is parallel to the field, the normal is perpendicular to B, so θ = 90°, cos(θ) = 0. Therefore Φ_initial = B A cos(90°) = 0.0 Wb.
- Finally, when the plane is perpendicular to the field, the normal aligns with B, so θ = 0°, cos(θ) = 1. Therefore Φ_final = B A cos(0°) = B A = 0.75 T × 0.40 m^2 = 0.30 Wb.

The change in magnetic flux through one turn is ΔΦ_one = Φ_final − Φ_initial = 0.30 − 0 = 0.30 Wb. For a coil with N turns, the total change in flux linkage is ΔΦ_total = N ΔΦ_one = 15 × 0.30 = 4.5 Wb.

The average induced emf in a coil is given by Faraday’s law: ε_avg = |ΔΦ_total| / Δt. Substituting the numbers: ε_avg = 4.5 Wb / 0.050 s = 90 V.

Now we examine the answer choices:
- 45 V: This would correspond to roughly half the flux change or half the turns contributing, which is not consistent with the full flux change of all 15 turns over the given time. It neglects the N factor and uses an incorrect flux value.
- 6.0 V: This is far too small for the given flux change and number of turns; it might arise from dividing by a larger Δt or forgetting the N factor, but it does not match the computed ΔΦ_total/Δt.
- 90 V: This exactly matches the calculation ε_avg = (N × ΔΦ_one) / Δt = (15 × 0.30 Wb) / 0.050 s = 90 V. It correctly accounts for both the flux change and the number of turns and the time interval.
- 36 V: This would imply a different combination, such as using a smaller ΔΦ or more modest factors, but does not align with the full flux linkage change over the interval for all turns.

In summary, the key steps are identifying the initial and final flux, computing the total flux linkage change across all turns, and dividing by the flipping time. Only the 90 V option is consistent with these steps, while the other options misrepresent either the flux, the number of turns, or the time interval.

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