Question

When the current in a toroidal solenoid is changing at a rate of 0.0200 A/s, the magnitude of the induced emf is 12.7 mV. When the current equals 1.50 A, the average flux through each turn of the solenoid is 0.00458 Wb. How many turns does the solenoid have?

Answer 1

Answer:

$N = 208 \ turns$

Explanation:

From the question we are told that

    The  rate of  current change is  $\frac{di }{dt} = 0.0200 \ A/s$

    The  magnitude of the induced emf is  $\epsilon = 12.7 \ mV = 12.7 *10^{-3} \ V$

     The  current is  $I = 1.50 \ A$

      The  average  flux is  $\phi = 0.00458 \ Wb$

Generally the number of  turns the number of turn the solenoid has is mathematically represented as  

            $N = \frac{\epsilon_o * I}{ \phi * \frac{di}{dt} }$

substituting values

           $N = \frac{ 12.7*10^{-3} * 1.50 }{ 0.00458 * 0.0200 }$

            $N = 208 \ turns$