How many windings needed to generate Peak emf

[musiliu]

Homework Statement

A simple generator has a coil with dimensions 1.0 cm x 1.0 cm which
is rotated at 60 Hz in 1.0 T magnetic field. How many windings (turns of
wire) are needed to generate a peak emf of 120 Volts ?

Homework Equations

1. emf E = - d (magnetic flux) / dt
2. magnetic flux = (B dot n hat) A

3. emf E for generator(rotating coil/loop) = N B A w(sin wt) where N is number of turns, B is magnetic field, A is area of coil, w is omega = 2 pi f

The Attempt at a Solution

Is it correct to assume the coil is a square coil? since it only gives me those two dimensions..

i used equation 3 above and used A = .01 m(squared), B = 1.0 T , omega w = 2 pi f, f = 60 Hz, and emf E = 120 V

but i don't know what "t" is for sin(wt)... do i even need that part of the equation to solve this problem?
what exactly does "peak emf" mean? does it mean amplitude, in which case i only need NBAw?

 

[Chi Meson]

Is it correct to assume the coil is a square coil? since it only gives me those two dimensions..

yes that is a safe assumption.

The peak emf occurs at the point where the change in flux is maximum. THis is the point where the coil is "sideways" to the magnetic field. The actual flux through the coil at this point is zero, but the "rate of change of flux" is what matters, and as the coil "crosses to the other side," the change in flux is greatest . Either way, you are looking for the point where emf is maximum, and that will be the point where sinwt is maximum, and the maximum sine for anything is 1.0, isn't it? So your assumption is correct.

 

[robcowlam]

Ok you're not far off with what you have got:
Assume the coil is square, this gives an area of 0.0001[m^2]
Now from equation 2 you said flux = B.A
as the coil is rotating the area of the coil which is perpendicular to the magnetic flux is actually given by flux=BACos(ωt).
and from 1 emf = -N.d(flux)/dt
now differentiating w.r.t t we get: -NBA d(cos(ωt))/dt = NBAωsin(ωt)
so EMF = NBAωsin(ωt)
you are asked for the maximum value of the emf to be 120v so from the sine curve we know the maximum value it can be is 1. (when ωt = 90, 270 etc)
so EMF(max)=NBAω
ω=2*pi*frequency
rearranging gives:
N=EMF/BA*2*pi*f

Hope that helps.

 

[musiliu]

ok, thanks, the max sin = 1 helped a lot...

so i found the answer to be N = 3183 turns

is this correct?

 

[rwisz]

or does it mean the maximum voltage output of the generator?
To generate a peak emf of 120 Volts, we can use the equation E = NBAw, where E is the emf, N is the number of turns, B is the magnetic field, A is the area of the coil, and w is the angular velocity. In this case, the angular velocity can be calculated using the frequency, f, which is given as 60 Hz. Therefore, w = 2pi*f = 2pi*60 = 120pi rad/s.

Since the dimensions of the coil are given as 1.0 cm x 1.0 cm, we can assume it is a square coil with an area of 0.01 m^2. Substituting these values into the equation, we get:

120 V = N * 1.0 T * 0.01 m^2 * 120pi rad/s

Solving for N, we get N = 1000 turns.

Therefore, 1000 turns of wire are needed to generate a peak emf of 120 Volts in this simple generator. The "peak emf" refers to the maximum voltage output of the generator, which occurs when the coil is in a position perpendicular to the magnetic field, maximizing the change in magnetic flux and inducing the highest emf.