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andriy [413]
3 years ago
8

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
Physics
1 answer:
GREYUIT [131]3 years ago
5 0

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

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