# Emf in AC generator?

pivoxa15

## Homework Statement

The armature of an AC generator is rotating at a constant speed of 30 revolutions/second in a horizontal field of flux density 1.0Wb/m^2. The diameter of the cylindrical armature is 24cm and its length is 40cm. What is the maximum emf induced in the armature having 30 turns?

## Homework Equations

emf=-magnetic flux rate although this problem is more about magnitude

## The Attempt at a Solution

The generator is rotating at 0.033seconds/period. The area facing the magnetic field follows a sin curve. The full area is (0.12m)^2*pi=0.045m^2 The sine curve has angular frequency b given by 2*pi/b=0.033seconds/period. So b=188.5 Hence the area is given by 0.045m^2*sin(188.5t)

The magnetic flux rate is dA/dt * B

dA/dt =0.045m^2*188.5*cos(188.5t)=8.48*cos(188.5t)

max magnetic flux rate = 8.48*30 = 254.5V which includes the 30 turns. The answer stated 543V. Why?
When doing the problem I have not used the length of the cylindrical armature which stands at 0.4m. I don't know where I could use it.

Homework Helper

## Homework Statement

The armature of an AC generator is rotating at a constant speed of 30 revolutions/second in a horizontal field of flux density 1.0Wb/m^2. The diameter of the cylindrical armature is 24cm and its length is 40cm. What is the maximum emf induced in the armature having 30 turns?

## Homework Equations

emf=-magnetic flux rate although this problem is more about magnitude

## The Attempt at a Solution

The generator is rotating at 0.033seconds/period. The area facing the magnetic field follows a sin curve. The full area is (0.12m)^2*pi=0.045m^2 The sine curve has angular frequency b given by 2*pi/b=0.033seconds/period. So b=188.5 Hence the area is given by 0.045m^2*sin(188.5t)

The magnetic flux rate is dA/dt * B

dA/dt =0.045m^2*188.5*cos(188.5t)=8.48*cos(188.5t)

max magnetic flux rate = 8.48*30 = 254.5V which includes the 30 turns. The answer stated 543V. Why?
When doing the problem I have not used the length of the cylindrical armature which stands at 0.4m. I don't know where I could use it.
The cylindrical armature is spinning on its long axis and the wire is wrapped in square loops around its length each measuring 24 x 40 cm. So each loop has an area of .24x.4 = .096 m^2.

AM

pivoxa15
So the wires are not wrapped around in circles but in rectangles with an area vector perpendicular to the 2 circlular faces of the cylinder. In fact the area vector of the 2 circular faces act as the axis of rotation and the area vector of the rectangle formed by the wires spin in a circlee and come back to the same position every 360 degrees.