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 9.50 ?. The coil rotates in a 0.520 T magnetic field at a constant frequency of 60.0 Hz. Use the following equation to find the maximum induced emf: emfmax = NAB? = NAB(2?f ) Substitute numerical values: emfmax = 8(0.0900 m2)(0.520 T)(2?)(60.0 Hz) emfmax = 141.14. V Imax = emfmax R = emfmax 9.50 ? Imax = 14.856. A ---My question is, what if at a later time, the maximum induced current was measured to be 6.50 A instead. Assuming the magnetic field is unchanged, at what frequency is the loop spinning? f = Hz

Explanation / Answer

I(max)=6.5 I(max)=v(max)/R v(max)=6.5*9.5=61.75 v(max)=NAB*2*pi*f f=26.26 hz

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---