The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radius of 1.20 m.

Question

The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radius of 1.20 m. The toroid has 900 turns of large diameter wire, each of which carries a current of 13.0 kA. Find the difference in magnitudes of the magnetic fields of the toroid along the inner and outer radii. (Enter your answer in T.)

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Thanh Thu 3 weeks 2021-08-22T18:32:39+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-22T18:34:37+00:00

    Answer:

    The difference is \Delta B  = 1.39 \ T

    Explanation:

    From the question we are told that  

        The inner radius  is  r_i  =  0.700 \ m

            The outer radius  is  r_o  = 1.20 \ m

            The number of turns is  N  = 900 \ turns

            The current on each wire is I  =  13.0 kA =  13*10^{3} \ A

    Generally magnetic field of a toroid along the outer radius is mathematically evaluated as

            B_o  =  \frac{\mu_o * N * I}{2 \pi r_o}

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

    substituting values

                B_o  =  \frac{ 4\pi * 10^{-7} *  13*10^{3} *  900}{ 2 * 3.142 * 1.20}

               B_o  =  1.95 \ T

    Generally magnetic field of a toroid along the inner radius is mathematically evaluated as

               B_i  =  \frac{\mu_o * N * I}{2 \pi r_i}

    substituting values

               B_i  =  \frac{ 4\pi * 10^{-7}   * 900 * 13*10^{3}}{2 *3.142 *0.700}

             B_i  = 3.34 \ T

    The difference in  magnitudes of the magnetic fields of the toroid along the inner and outer radii is mathematically evaluated as

          \Delta B  =  B_i - B_o

          \Delta B  = 3.34 -1.95

          \Delta B  = 1.39 \ T

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