Question

a simple generator has 200 loop square coil 10 cm on a side .how fast must it turn in a 0.25 T field to produce 24v peak out put​

Answer 1

Hi there!

Recall Faraday's Law:
$\epsilon = N\frac{d\Phi_B}{dt}$

ε = Emf (V)
N = Number of loops

ΦB = Magnetic Flux (Wb)
t = time (s)

Since the magnetic field is constant, we can take this out of the time derivative:
$\frac{d\Phi_B}{dt} = B * \frac{dA}{dt}$

Therefore:
$\epsilon = N B \frac{dA}{dt}$

We can express 'A', the area in which the magnetic field passes as:

$A = Acos(\omega t)$

Taking the time derivative:

$\frac{dA}{dt} = A\omega sin(\omega t)$

ω = angular speed of coil (rad/sec)

Now, combine with the above expression:

$\epsilon = NBA\omega sin(\omega t)$

The maximum output will occur when the loop's area vector is PERPENDICULAR to the field, so sin(ωt) = 1.

Therefore:
$\epsilon = NBA\omega$

Rearrange to solve for ω:

$\omega = \frac{\epsilon}{NBA}\\\\\omega = \frac{24}{(200 * (.10^2) * 0.25} = \boxed{48 \frac{rad}{sec}}$