# A 9-turn generator coil with area, 0.160m^2 rotates in a 1.30T magnetic field. At what frequency...

## Question:

A 9-turn generator coil with area, 0.160m{eq}^2 {/eq} rotates in a 1.30T magnetic field. At what frequency should the coil rotate to produce a peak emf of 135, as expressed using three significant figures?

## Faraday's Law of Electromagnetic Induction:

When there is a change in the magnetic flux passing through a coil, an EMF is induced in the coil. The magnitude of the induced EMF is equal to the rate of change of magnetic flux.

{eq}\begin{align*} \phi &= N\overrightarrow{B}.\overrightarrow{A} \\ \epsilon &=- \frac{d \phi}{dt}\\ \end{align*}{/eq}

Where:

• {eq}\phi{/eq} is the flux passing through N turns of the coil having area A
• B is the magnetic field in which the coil is kept
• {eq}\epsilon{/eq} is the induced EMF in the coil

NOTE: The negative sign gives us the polarity of the induced EMF.

We are given:

• Number of turns of the generator coil is {eq}N = 9 {/eq}
• Magnetic field in the region is {eq}B =1.30 \ T{/eq}
• Area of the coil having dimensions is {eq}A = 0.160 \ m^2{/eq}
• The maximum induced EMF of the coil is {eq}\varepsilon_{max} = 135 \ V {/eq}.

The angular deviation of rotation is given by:

{eq}\theta = \omega t{/eq}

From the above expression of flux, we have:

{eq}\begin{align*} \phi (t) &= N\overrightarrow{B}.\overrightarrow{A} \ \Rightarrow Dot \ product \\ &=NBA \cos \theta \ \Rightarrow where \ \theta = angular \ rotation \ of \ the \ coil \\ \boxed {\phi (t) =NBA \cos \omega t} \end{align*}{/eq}

To calculate the induced EMF, we have:

{eq}\begin{align*} \epsilon &=- \frac{d \phi}{dt}\\ &=- \frac{d (NBA \cos \omega t )}{dt}\\ &=NBA \omega \sin \omega t \\ \boxed { \epsilon=NBA \omega \sin \omega t} \end{align*}{/eq}

Now to produce maximum induced EMF, we have:

{eq}sin \theta = max{/eq}

In other words:

{eq}\theta = 90 ^o{/eq}

{eq}\begin{align*} \varepsilon_{max} &=NBA \omega \sin \theta \\ 135 &= 9 \times 1.30 \times0.160 \times \omega \times \sin 90^o \\ \omega & =72.1 \ rad/s\\ 2\pi f & = 72.1 \\ \boxed {Frequency = f =11.5 \ Hz } \Rightarrow(Answer) \end{align*}{/eq} 