Question

Show that if a coil rotates at an angular velocity \({\rm{\omega }}\)the period of its AC output is$\frac{2\pi }{\omega }$

Short answer

The time period of AC output with angular velocity $\omega$ is $\frac{2\pi }{\omega }$.

Step-by-step solution

Step 1:Explanation

Angular velocity can be defined as the change that occurs in the angular position in unit time. The time taken by the coil to complete a rotation is called Time Period.

Explanation for the time period of AC output with angular velocity \({\rm{\omega }}\)

The output of an AC generator is given by

$\epsilon \left(t\right)=NAB\sin \left(\omega t\right)$

Where $\omega$is the angular frequency of the generator, N is the number of turns in the coil, B is the magnetic field applied, $\epsilon$.is the emf produced, A is the cross-sectional area and t is the time.

Even though it might seem trivial, let us say that since the aforementioned relation holds true, we can see that the angular frequency of the generated AC is the same as that of the generator's shaft (and not, for example, a multiple of it).

Keeping the notation the same,

$$\begin{gathered} {\omega }_{gen}={\omega }_{AC} \\ =\omega \end{gathered}$$

The relation between the angular velocity and time period for the rotation fo the coil, is-

$\omega T=2\pi rad/s$

From this, we can find that,

$T=\frac{2\pi }{\omega }$

Therefore, the time period of AC output with angular velocity$\omega$ is $\frac{2\pi }{\omega }$.