The coil in an AC generator consists of 8 turns of wire, each of area A = 0.0900

Question

The coil in an AC generator consists of 8 turns of wire, each of area A = 0.0900 m2, and the total resistance of the wire is 12.0 ?. The coil rotates in a 0.610 T magnetic field at a constant frequency of 60.0 Hz. Use the following equation to find the maximum induced emf: emfmax = NAB(2pi f ) Substitute numerical values: emfmax = 9(0.0900 m2)(0.610 T)(2pi)(60.0 Hz). emfmax = 186.27 V. Imax = emfmax/ R = emfmax/12.0 ?. Imax = 15.5 A ---My question is, what if at a later time, the maximum induced current was measured to be 6.00 A instead. Assuming the magnetic field is unchanged, at what frequency is the loop spinning? f = Hz

Explanation / Answer

here ,

for Imax = 6 A

as I max is proporational to frequency ,

for other values constant

I1/I2 = f1/f2

6/15.5 = f1/60

f1 = 23.22 hz

the new frequency of loop is 23.22 Hz

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