High School

A solenoid with 1,107 turns has a radius of 82.4 mm and is 39.4 cm long. If this solenoid carries a current of 67.2 A, what is the magnitude of the magnetic field near the center of the solenoid?

Answer :

Final answer:

The magnitude of the magnetic field near the center of the solenoid is approximately 0.670 T.

Explanation:

To determine the magnitude of the magnetic field near the center of a solenoid, you can use the formula:

B = μ₀ * N * I / L

Where:

  • B is the magnetic field strength
  • μ₀ is the permeability of free space (4π × 10-7 T m/A)
  • N is the number of turns in the solenoid
  • I is the current through the solenoid
  • L is the length of the solenoid

Plugging in the given values:

  • μ₀ = 4π × 10-7 T m/A
  • N = 1,107 turns
  • I = 67.2 A
  • L = 0.394 m

Calculating:

  1. B = (4π × 10-7) * (1,107) * (67.2) / (0.394)
  2. B ≈ 0.670 T

Therefore, the magnitude of the magnetic field near the center of the solenoid is approximately 0.670 T.

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