Question

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)

Answer 1

Answer:

The induced  emf is  $\epsilon =7.68 \ V$

Explanation:

From the question we are told that

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

    The diameter is  $d = 2.2 \ cm = 2.2*10^{-2}$

     The initial magnetic field is  $B_i = 0.14 \ T$

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

      The  time taken is  $dt = 9.0ms = 9.0*10^{-3} \ s$

 

The radius is mathematically evaluated as

      $r = \frac{d}{2 }$

substituting values

     $r = \frac{2.2 *10^{-2}}{2 }$

     $r = 1.1*10^{-2} \ m$

The induced emf is mathematically represented as

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

Where  $d\phi$ is the change in magnetic field which is mathematically represented as

        $d\phi = dB * A * cos\theta$

=>   $d\phi = [B_f - B_i ] * A * cos\theta$

Here  $\theta = 0$ given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

       $A = \pi r^2$

substituting values

      $A = 3.142 * [1.1*10^{-2}]^2$

       $A = 3.8 *10^{-4] \ m^2$

So

    $d\phi = [0 - 0.14 ] * 3.8*10^{-4}$

    $d\phi = -5.32*10^{-5} \ weber$

So  

     $\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }$

    $\epsilon =7.68 \ V$