Question

A 6.7 cm diameter circular loop of wire is in a 1.27 T magnetic field. The loop is removed from the field in 0.16 ss . Assume that the loop is perpendicular to the magnetic field.

Required:
What is the average induced emf?

Answer 1

Answer:

The induced emf is $\epsilon = 0.0280 \ V$

Explanation:

From the question we are told

    The diameter of the loop is  $d = 6.7 cm = 0.067 \ m$

    The magnetic field is  $B = 1.27 \ T$

    The time taken is  $dt = 0.16 \ s$

Generally the induced emf is mathematically represented as

          $\epsilon = - N * \frac{\Delta \phi}{dt}$

Where  N =  1 given that it is only a circular loop

            $\Delta \phi = \Delta B * A$

Where  $\Delta B = B_f - B_i$

   where $B_i$ is  1.27 T  given that the loop of wire was initially in the magnetic field

    and  $B_f$ is  0 T given that the loop was removed from the magnetic field

Now the area of the of the loop is evaluated as

          $A = \pi r^2$

Where r is the radius which is mathematically represented as

       $r = \frac{d}{2}$

substituting values

       $r = \frac{0.067}{2}$

        $r = 0.0335 \ m$

So

         $A = 3.142 * (0.0335)^2$

          $A = 0.00353 \ m^2$

So

      $\Delta \phi = (0- 127)* (0.00353)$

      $\Delta \phi = -0.00448 Weber$

=>    $\epsilon = - 1 * \frac{-0.00448}{0.16}$

=>   $\epsilon = 0.0280 \ V$