Question

Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.012 m, and carries a current of 6.2 A. The outer coil contains 220 turns and has a radius of 0.020 m. What must be the magnitude of the current in the outer coil, such that the net magnetic field at the common center of the two coils is zero

Answer 1

Answer:

The current of the outer coil is  $I_o = 3.99 \ A$

Explanation:

From the question we are told that

    The number of turns of the inner coil is  $N_i = 170 \ turn$

    The radius of the  inner coil is $R_i = 0.012 \ m$

     The current of the inner coil is  $I_i = 6.2 \ A$

      The number of turns of the outer coil is $N_o = 220 \ turns$

      The radius of the  outer coil is $R_o = 0.02 0 \ m$

       

Generally the net magnetic field is mathematically represented as

              $B = \frac{N \ mu I }{ 2 * R }$

Now from told that the net magnetic field is common

So  

           $\frac{N_i \mu I_i} {2 * R _i} = \frac{N_o \mu I_o} {2 * R _o}$

Here  $\mu$ is the permeability of free space

making  $I_o$ the subject

            $I_o = \frac{ N_i I_i *2 * R _o}{N_o *2 * R _i}$

substituting values

           $I_o = \frac{ 170 *6.2 *2 * 0.020}{220 *2 * 0.012}$

         $I_o = 3.99 \ A$