Question

A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.A) What is the magnetic field in the core? B) What part of the magnetic field is due to atomic currents?

Answer 1

Answer

given,

number of turns, N = 400 turns

mean radius, R = 6 cm

Current, I = 0.25 A

relative permeability, K_m = 80

a) Magnetic field in the core

  $B =\dfrac{K_m\mu_o N I}{2\pi R}$

  $B =\dfrac{80\times 4\pi \times 10^{7}\times 400 \times 0.25}{2\pi\times 0.06}$

        B = 2.67 x 10⁻² T

b) the amount of magnetic field due to atomic current

    $B = \dfrac{K_m\mu_o N I}{2\pi R} - \dfrac{\mu_o N I}{2\pi R}$

    $B = \dfrac{(K_m-1)\mu_o N I}{2\pi R}$

    $B = \dfrac{(80-1)\times 4\pi \times 10^{7}\times 400 \times 0.25}{2\pi\times 0.06}$

         B = 2.63 x 10⁻² T