Two circular coils are concentric and lie in the same plane. The inner coil contains 110 turns of wire, has a radius of 0.014 m, and carries

Question

Two circular coils are concentric and lie in the same plane. The inner coil contains 110 turns of wire, has a radius of 0.014 m, and carries a current of 9.0 A. The outer coil contains 160 turns and has a radius of 0.022 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?

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Edana Edana 5 years 2021-09-02T12:48:01+00:00 1 Answers 31 views 0

Answers ( )

  1. Ben
    0
    2021-09-02T12:49:33+00:00
    Reply

    Answer:

    The current flowing through the outer coils is  

    Explanation:

    From the question we are told that

       The number of turn of inner coil is N _i = 110 \ turns

        The radius of inner coil is  r_i = 0.014 \ m

         The current flowing through the inner coil is  I_i = 9.0 \ A

         The number of turn of outer coil is N_o = 160 \ turns

         The radius of outer  coil is r_o = 0.022\ m

    For net magnetic field at the common center of the two coils to be  zero  the current flowing in the outer coil must be opposite to current flowing inner coil

       The magnetic field due to inner coils  is mathematically represented as

                B_i = \frac{N_i \mu I}{2 r_i}

         The magnetic field due to inner coils  is mathematically represented as

                B_o = \frac{N_o \mu I_o}{2 r_o}

    Now for magnetic field at center to be zero

                 B_o = B_i

    So

             \frac{N_i \mu I_i}{2 r_i} = \frac{N_o \mu I_o}{2 r_o}

    =>      \frac{110 * 9}{2 * 0.014} = \frac{160 *I_o}{2 0.022}

             I_o = 9.72 \ A

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