Question

The current in a long solenoid of radius 5 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of radius 7 cm and resistance 5 Ω surrounds the solenoid. Find the electrical current induced in the loop (in µA).

Answer 1

Answer:

The current is  $I =0.2mA$

Explanation:

From the question we are told that

   The first radius is $R_1 = 5cm = \frac{5}{100} = 0.05cm$

    The number of turns is $N = 17 \ turn/cm$

    The current rate is  $\frac{dI}{dt} = 5 A/s$

    The second radius is  $R_2 = 7cm = \frac{7}{100} = 0.07m$

     The resistance is $r = 5 \Omega$

Generally the magnetic flux induced in the solenoid is mathematically represented as

      $\O = B A$

 Where  is the magnetic field mathematically represented as

            $B = N \mu_o I$

Where $\mu_o$ is the permeability of free space with a value of $\mu_o = 4\pi *10^{-7} N/A^2$

     and A is the area mathematically represented as

             $A = \pi (R_2 - R_1)^2$

So

         $\O = N \mu I * \pi R^2$

            Substituting values

        $\O = 17 * 4\pi *10^{-7} * \pi (7-5)^2I$

           $\O = 2.68*10^{-4}I$

The induced emf is mathematically represented as

              $\epsilon =- |\frac{d\O}{dt}|$

                  $\epsilon = 2.68*10^{-4 } \frac{dI}{dt}$

substituting values

               $\epsilon =2.68 *10^{-4} * 5$

                  $=1.3 *10^{-3} V$

From Ohm law

      $I = \frac{\epsilon }{r}$

Substituting values

     $I = \frac{1.3*0^{-3}}{5}$

        $I =0.2mA$