the electric generator

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

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 = [itex] n B.A [/itex]
where n is the number of loops, B is the magnetic field and A is the area of the loop.

[tex] = (n)(B)(A)(\cos\theta) [/tex]

So, the EMF induced will be

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

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)

KingOfTwilight

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

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

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

Just find [tex] \frac{dA}{dt}[/tex] and the biggest problem is probably solved.

mayo2kett

thanks guys... you really helped me