Respuesta :

onyebuchinnaji

The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].

The given parameters;

  • length of the solenoid, L = 91 cm = 0.91 m
  • radius of the solenoid, r = 1.5 cm = 0.015 m
  • number of turns of the solenoid, N = 1300
  • current in the solenoid, I = 3.6 A

The magnitude of the magnetic field inside the solenoid is calculated as;

[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]

where;

[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A

[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]

Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].

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