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Magnetization. Consider a long solenoid with a core that is an iron alloy. Suppose that the diameter of the solenoid is 2 cm and the length of the solenoid is 20 cm. The number of turns on the solenoid is 200. The current is increased until the core is magnetized to saturation at about I= 2 A and the saturated magnetic field is 1.5 T.

(a) What is the magnetic field intensity at the center of the solenoid and the applied magnetic field, , for saturation?

User MmBs
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Final answer:

The magnetic field intensity at the center of a solenoid with an iron alloy core is determined by the number of turns per unit length, current, and permeability of free space. At saturation, with a given current, the magnetic field is approximately 1.5 T. The applied magnetic field can be calculated using the turns per unit length and current.

Step-by-step explanation:

Magnetic Field Intensity and Applied Magnetic Field in a Saturated Solenoid with an Iron Alloy Core

To determine the magnetic field intensity at the center of a solenoid and the applied magnetic field at saturation, we use the formula for the magnetic field inside a solenoid, which is given by:

B = μ₀ × n × I

where:

  • B is the magnetic field intensity,
  • μ₀ is the permeability of free space (4π × 10⁻⁷ T⋅m/A),
  • n is the number of turns per unit length of the solenoid (×200/0.2 = 1000 turns/m),
  • I is the current (2 A).

Using the given dimensions and current, the magnetic field intensity at the center of the solenoid at saturation can be calculated. As the solenoid's core is an iron alloy and it is saturated, the magnetic field would be approximately 1.5 T as stated, which represents the maximum level of magnetization the material can achieve.

The applied magnetic field (H) for saturation can also be calculated using:

H = n × I

where H is the applied magnetic field strength. We already know n and I, so we can compute H accordingly.

User Stasiaks
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