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a 38.5 ma current is carried by a uniformly wound air-core solenoid with 460 turns, a 19.0 mm diameter, and 10.5 cm length.(a) Compute the magnetic field inside the solenoid. �T (b) Compute the magnetic flux through each turn. T�m2 (c) Compute the inductance of the solenoid. mH(d) Which of these quantities depends on the current? (Select all that apply.) magnetic field inside the solenoid magnetic flux through each turn inductance of the solenoid

Respuesta :

The magnetic field inside the solenoid is 1.88 × 10^-4 T.

The magnetic flux through each turn is 330 × 10^-10 Tm²

I = 38.5 × 10^-3 A

N = 460 turns

d= 19 mm = 19 × 10^-3 m

r= 9.5 × 10^-3 m

L = 10.5 × 10^-2 m

Hence,

Magnetic field = μ0nI

=(4π/10⁷)(460/10.5×10^-2)(38.5×10^-3)

B= 1.88 × 10^-4 T

Magnetic flux= BA = 330 × 10^-10 Tm²

What is magnetic field?

  • Magnetic fields are created by moving electric charges and intrinsic magnetic moments of elementary particles associated with a basic quantum property, their spin.
  • Magnetic fields and electric fields are interrelated and both are part of the electromagnetic force, one of the four fundamental forces of nature.

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