Question

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 3.5 mV/m. At what rate is the magnetic field changing?

Answer 1

Answer

The rate at which the magnetic field is changing is  $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

   The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

    The radius is  $r = 1.5 \ m$

The rate of change of the  magnetic  field  is mathematically represented as

        $\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

     $\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

     $A = \pi r^2$

    So

    $E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$  

  $E L = ( \pi r^2) (\frac{dB}{dt} )$  

where L is the circumference of the circle which is mathematically represented as

     $L = 2 \pi r$

So

     $E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

      $E = \frac{r}{2} [\frac{dB}{dt} ]$

       $[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

      $[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

      $[\frac{dB}{dt} ] = 0.000467 T/s$