# A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 5.00 cm by 6.40 cm...

## Question:

A car generator turns at 400 rpm when the engine is idling. Its 300-turn, 5.00 cm by 6.40 cm rectangular coil rotates in an adjustable magnetic field so that it can produce sufficient voltage even at low rpms.

What is the field strength needed to produce a 24.0 V peak emf?

## Generator

Electric generator is a device which converts mechanical energy to electrical energy.

For a AC generator, the peak induced emf is related to the area of the coil (A), number of turns of the coil (N) and magnetic field (B) as, {eq}\varepsilon _{peak} = NBA \omega {/eq}, where

• {eq}\omega {/eq} is the angular speed of rotation of coil.

## Answer and Explanation:

Given :

The angular speed of the coils is, {eq}\omega = 400 \ rpm = 400 \times \dfrac{2 \pi}{60} \ rad/s = 41.888 \ rad/s {/eq}

The area of the coil is, {eq}A = (5 \times 6.4) \ cm^2= 32 \ cm^2 = 0.0032 \ m^2 {/eq}

The number of turns on the coil are, {eq}N = 300 {/eq}

The peak emf induced in the coil is, {eq}\varepsilon _{peak} = 24 \ V {/eq}

Let the strength magnetic field be, B

The peak induced emf in the generator is given by, \begin{align*} \varepsilon _{peak} &= NBA \omega \\ 24 &= (300)(B)(0.0032) (41.888 ) \\ B &= 0.597 \ T \end{align*}