Question

A generator can be made using the component of Earth’s magnetic field that is parallel to Earth’s surface. A 140-turn square wire coil with an area of 4.30 ×10−2 m2 is mounted on a shaft so that the cross-sectional area of the coil is perpendicular to the ground. The shaft then rotates with a frequency of 21.4 Hz. The horizontal component of the Earth’s magnetic field at the location of the loop is 3.00 ×10−5 T.
Calculate the maximum emf induced in the coil by Earth's magnetic field.

Answer 1

Answer:

The maximum induced emf is    $\epsilon = 0.024 V$

Explanation:

From the question we are told that

   The number of turns is $N = 140 \ turns$

    The area of the square wire coil is  $A = 4.30 * 10^{-2}m^2$

    The frequency of rotation is  $f = 21.4 Hz$

    The earth magnetic field at that location is  $B = 3.00 *10^{-5} T$

The  induced emf is mathematically represented as

         $\epsilon = N \ B \ A \ w sin(wt)$

At the maximum

       $sin (wt) = 1$

So

         $\epsilon = N \ B \ A \ w$

Where

    $w$ is the angular velocity which is mathematically represented as

          $w =2 \pi f$

Substituting values

             $w = 2 * 3.142 * 21.4$

            $w = 134.5 rad/sec$

The maximum induced emf is  

          $\epsilon = 140 * 3.00*10^{-5} * (4.30 *10^{-2}) * (134.5)$

           $\epsilon = 0.024 V$