# The electric generator

1. Jul 29, 2005

### mayo2kett

The coil of a generator has a radius of 0.14 m. When this coil is unwound, the wire from which it is made has a length of 5.4 m. The magnetic field of the generator is 0.10 T, and the coil rotates at an angular speed of 35 rad/s. What is the peak emf of this generator?

so i have:
r= .14m
L= 5.4m
B= .10T
w=35 rad/s

now i thought i would do:
emf= BLv
v=rw... v= .14m(35 rad/s)
emf= (.10T)(5.4m)(.049m/s)= .02646
and peak emf= (square root 2)(emf)= .0374...

this problem is wrong the way i tried it, but i'm not sure what i should do differently

2. Jul 30, 2005

### siddharth

The induced EMF (across the ends of the rod) due to the motion of a rod of length 'l' and velocity 'v', in the presence of a magnetic field of strength 'B' is Blv. So this formula is not applicable here as there is a rotating coil and not a rod.

To solve this problem, go from the definition of Farady's law.
By Farady's law, Emf induced = -d(Magnetic Flux)/dt

Let the magnetic field make an angle theta with the area vector of the loop at any time 't' such that at t=0, theta=0.
So the Magnetic flux enclosed by the loop is = $n B.A$
where n is the number of loops, B is the magnetic field and A is the area of the loop.

$$= (n)(B)(A)(\cos\theta)$$

So, the EMF induced will be

$$=\frac {-d[(n)(B)(A)(\cos\theta)]}{dt}$$

From this, can you calculate the EMF as a function of time and from that the peak value?
(You will have to find the relation between 'theta' and 't' as well as the value of n)

Last edited: Jul 30, 2005
3. Jul 30, 2005

### KingOfTwilight

The coil is rotating in the field. The flux is thus changing and this causes the electric field in the coil.

$$\Phi = AB$$, B is constant but A is changing. Can you find A as a function of time?

$$E = -N \frac{d\Phi}{dt}$$, so you will also need to find N - the number of layers in the coil.

Just find $$\frac{dA}{dt}$$ and the biggest problem is probably solved.

4. Jul 30, 2005

### mayo2kett

thanks guys... you really helped me

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