Question

A 1300-turn coil of wire that is 2.2 cm in diameter is in a magnetic field that drops from 0.11 T to 0 {\rm T} in 12 ms. The axis of the coil is parallel to the field What is the emf of the coil?

Answer 1

Answer:

The induced voltage is  $\epsilon = 4.53 \ V$

Explanation:

From the question we are told that

    The  number of turns is  $N = 1300 \ turns$

     The diameter is  $d = 2.2 \ cm =0.022 \ m$

     The  initial magnetic field is  $B_i = 0.11 \ T$

     The final magnetic field is  $B_f = 0 \ T$

    The time taken is  $t = 12 \ ms = 12*10^{-3} \ s$

   

The radius is mathematically evaluated as

         $r = \frac{d}{2}$

substituting values

         $r = \frac{0.022}{2}$

         $r = 0.011 \ m$

Generally the induced emf  is mathematically represented as

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

Where  $d\phi$ is the change in magnetic flux of the wire which is mathematically represented as

      $d \phi = dB* A * cos \theta$

=>  $d \phi = (B_f - B_i )* A * cos \theta$

Here  $\theta = 0$

since the axis of the coil is parallel to the field

    Where A  is the cross-sectional area of the coil which is mathematically represented as

      $A = \pi * r^2$

       $A = 3.142 * 0.011^2$

      $A = 3.80*10^{-4} \ m^2$

So the induced emf

        $\epsilon = - 1300 * \frac{(0- 0.11) * 3.80*10^{-4}}{12*10^{-3}}$   Here we substituted the values of  $d \phi$

       $\epsilon = 4.53 \ V$