The plane of a wire loop with an area of 0.20 m² is perpendicular to a magnetic field of 50 mT.In 40 ms you rotate the loop 90º so that its plane is parallel to the magnetic field.The average EMF induced in the loop is A)0.25 V B)6.0 V C)3.0 V D)4.0 V E)1.3 V

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Quizplus · Accepted Answer

The average EMF ($ \epsilon $) induced in the loop can be calculated using Faraday's law of electromagnetic induction, which states that $ \epsilon = -\Delta \Phi / \Delta t $, where $ \Delta \Phi $ is the change in magnetic flux and $ \Delta t $ is the time interval. Initially, the magnetic flux ($ \Phi $) is $ B \cdot A \cdot \cos(\theta) $, where $ B = 50 \, \text{mT} = 50 \times 10^{-3} \, \text{T} $, $ A = 0.20 \, \text{m}^2 $, and $ \theta = 0^\circ $ (since the plane of the loop is perpendicular to the magnetic field). When the loop is rotated $ 90^\circ $, $ \theta = 90^\circ $, and the plane of the loop is parallel to the magnetic field, resulting in $ \Phi = 0 $ because $ \cos(90^\circ) = 0 $. The change in flux $ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = 0 - (50 \times 10^{-3} \cdot 0.20 \cdot \cos(0)) = -0.01 \, \text{Wb} $. Given $ \Delta t = 40 \, \text{ms} = 40 \times 10^{-3} \, \text{s} $, the average EMF is $ \epsilon = -(-0.01) / (40 \times 10^{-3}) = 0.25 \, \text{V} $.

The Plane of a Wire Loop with an Area of 0.20