Question

Not yet answered Points out of 1.00 p Flag question What is the maximum Emf induced in a 918 turn coil square coil with sides of length 0.22 meters rotating at 568 rpm in a magnetic field of 0.068 Tesla? O a. 18.98 Volts

o b. 564.58 Volts O c. 816.87 Volts O d. 1716.11 Volts o e. 179.71 Volts

Answer 1

Using Faraday's law of electromagnetic induction, the maximum induced EMF is calculated as 179.71 volts, which is option e in the possible answers provided. Correct option 179.71 volts

Step-by-step explanation:

The maximum EMF induced in a coil rotating within a magnetic field can be calculated using Faraday's Law of electromagnetic induction, which states that the induced EMF is equal to the rate of change of magnetic flux, given by the equation: EMF = -N * (change in magnetic flux)/change in time, where N is the number of turns. In this case, we can use another form of the equation to accommodate the coil rotating in the magnetic field, which is EMF = NBAω, where N is the number of turns, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity of the coil. Substituting the given values:

• N = 918 turns
• B = 0.068 Tesla
• A = (0.22 m)^2 = 0.0484 m^2 (since the coil is square)
• ω = 568 rev/min * (2π rad/rev) * (1 min/60 s) = 59.5 rad/s

Substituting the values, we get EMF = 918 * 0.068 Tesla * 0.0484 m^2 * 59.5 rad/s = 179.71 Volts. Therefore, the maximum induced EMF is 179.71 volts, which corresponds to option e.

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