Question

A long, thin solenoid has 450 turns per meter and a radius of 1.17 cm. The current in the solenoid is increasing at a uniform rate did. The magnitude of the induced electric field at a point which is near the center of the solenoid and a distance of 3.45 cm from its axis is 8.20×10−6 V/m.
Calculate di/dt
di/dt = _________.

Answer 1

Answer:

$\frac{di}{dt} = 7.31 \ A/s$

Explanation:

From the question we are told that  

     The  number of turns is  $N = 450 \ turns$

      The  radius is  $r = 1.17 \ cm = 0.0117 \ m$

       The  position from the center consider is  x =  3.45 cm  =  0.0345 m

       The  induced emf is  $e = 8.20 *10^{-6} \ V/m$

Generally according to Gauss law

        $\int\limits { e } \, dl = \mu_o * N * \frac{di}{dt } * A$

=>    $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * A$

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

                $A = \pi r ^2$

=>      $e * 2\pi x = \mu_o * N * \frac{d i }{dt } * \pi r^2$

=>       $\frac{di}{dt} = \frac{2e * x }{\mu_o * N * r^2}$ggl;

Here  $\mu_o$ is the permeability of free space with value

          $\mu_o = 4\pi * 10^{-7} \ N/A^2$

=>     $\frac{di}{dt} = \frac{2 * 8.20*10^{-6} * 0.0345 }{ 4\pi * 10^{-7} * 450 * (0.0117)^2}$

=>      $\frac{di}{dt} = 7.31 \ A/s$