Calculating Current Needed for Solenoid

How to Calculate Current for a Solenoid

A 30.0-cm-long solenoid 1.25 cm in diameter is to produce a field of 4.65mT at its center. How much current should the solenoid carry if it has 935 turns of the wire?

The solenoid should carry approximately 1.17 A of current to produce a magnetic field of 4.65 mT at its center.

To find the current needed for a 30.0-cm-long solenoid with 1.25 cm in diameter to produce a field of 4.65 mT at its center and has 935 turns of wire, proceed as follows:

  1. First, we need to use the formula for the magnetic field B at the center of a solenoid: B = μ₀ * n * I, where μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), n is the number of turns per unit length (turns/m), and I is the current (A).
  2. Convert the length of the solenoid to meters: 30.0 cm = 0.3 m.
  3. Calculate the number of turns per unit length (n): n = total turns / length = 935 turns / 0.3 m = 3116.67 turns/m.
  4. Rearrange the formula for the magnetic field to solve for current: I = B / (μ₀ * n).
  5. Plug in the values for B, μ₀, and n: I = (4.65 × 10⁻³ T) / ((4π × 10⁻⁷ T·m/A) * 3116.67 turns/m).
  6. Calculate the current: I ≈ 1.17 A.
Question:

How much current should a 30.0-cm-long solenoid with 1.25 cm diameter and 935 turns of wire carry to produce a magnetic field of 4.65mT at its center?

Answer:

The solenoid should carry approximately 1.17 A of current to produce a magnetic field of 4.65 mT at its center. The calculation involves determining the number of turns per unit length, applying the formula for the magnetic field, and solving for current using the given values of magnetic field strength, permeability of free space, and number of turns.

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