Question

A very long, straight solenoid with a cross-sectional area of 2.22 cm2 is wound with 85.6 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)=(0.177A/s2)t2. A secondary winding of 5.0 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2A?

Answer 1

Answer:

The induce emf is  $\epsilon = 1.7966*10^{-5} V$

Explanation:

From the question we are told that

   The cross-sectional area is  $A = 2,22 cm^3 = \frac{2.22}{10000} = 2.22*10^{-4} \ m^2$

   The number of turn is $N = 85.6 \ turns/cm = 85.6 \ \frac{turns }{\frac{1}{100} } = 8560 \ turns / m$

   The starting time is  $t_o$ = 0 s

    The current increase is  $I(t) = (0.177A/s^2) t^2$

    The number of turn of secondary winding is  $N_s = 5 \ turn s$

     The current at the solenoid is   $I_(t) = 3.2 \ A$

at $I_(t) = 3.2 \ A$

         $3.2 = 0.177* t^2$

=>      $t = \sqrt{ \frac{3.2}{0.177} }$

        $t = 4.25 s$

Generally Faraday's law of induction is mathematically represented as

          $\epsilon = A\mu_o N_s N * \frac{di}{dt}$

         $\epsilon = A\mu_o N_s N * \frac{d (0.177 t^2)}{dt}$

          $\epsilon = A\mu_o N_s N * (0.177)(2t)$

substituting values

          $\epsilon = (2.22*10^{-4}) * ( 4\pi * 10^{-7}) * 5 * [8560]* 0.177 * 2 * 4.25$

          $\epsilon = 1.7966*10^{-5} V$