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 0.75-T $\vec { B }$ field.The coil is flipped so its plane is perpendicular to the $\vec { B }$ field in a time of 0.050 s.What is the magnitude of the average induced emf in the coil? A)6.0 V B)36 V C)45 V D)90 V

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Quizplus · Accepted Answer

The average induced emf ($ \epsilon $) can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the negative change in magnetic flux ($ \Delta \Phi_B $) over the change in time ($ \Delta t $). The magnetic flux ($ \Phi_B $) is given by $ \Phi_B = B \cdot A \cdot \cos(\theta) $, where $ B $ is the magnetic field strength, $ A $ is the area of the coil, and $ \theta $ is the angle between the magnetic field and the normal to the surface of the coil. Initially, the coil is parallel to the magnetic field ($ \theta = 0^\circ $), and finally, it is perpendicular ($ \theta = 90^\circ $), so the change in flux is from maximum to zero.Given:- Number of turns ($ N $) = 15- Area ($ A $) = 0.40 m$^2$- Magnetic field ($ B $) = 0.75 T- Time ($ \Delta t $) = 0.050 sInitial flux ($ \Phi_{B1} $) = $ B \cdot A $ (since $ \cos(0^\circ) = 1 $)Final flux ($ \Phi_{B2} $) = 0 (since $ \cos(90^\circ) = 0 $)Change in flux ($ \Delta \Phi_B $) = $ \Phi_{B2} - \Phi_{B1} = 0 - B \cdot A $ = $ -B \cdot A $Average induced emf ($ \epsilon $) = $ -N \cdot \frac{\Delta \Phi_B}{\Delta t} $Plugging in the values:$ \epsilon = -15 \cdot \frac{-0.75 \cdot 0.40}{0.050} $$ \epsilon = 15 \cdot \frac{0.30}{0.050} $$ \epsilon = 15 \cdot 6 $$ \epsilon = 90 $ VTherefore, the magnitude of the average induced emf in the coil is 90 V.

A Flat Square Coil of Wire with 15 Turns and an Area