Question

Your employer asks you to build a 26-cm-long solenoid with an interior field of 4.1 mT . The specifications call for a single layer of wire, wound with the coils as close together as possible. You have two spools of wire available. Wire with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A. Which wire should you use, and what current will you need?

Answer 1

To determine which wire to use for the solenoid, we need to calculate the current required. We can use the formula B = μ₀nI, where B is the magnetic field. Plugging in the given values, we can calculate the current that the wire can carry using the formula I = J x A. By comparing the calculated currents for both wires, you can determine which wire to use.

Step-by-step explanation:

To determine which wire to use for the solenoid, we need to calculate the current required. We can use the formula B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. Rearranging the formula to solve for I, we get I = B / (μ₀n). Plugging in the given values, we have I = 4.1 x 10^(-3) T / (4π x 10^(-7) Tm/A x 26 x 10^(-2) m).

The wire with #26 gauge has a diameter of 0.41 mm, which is equivalent to 4.1 x 10^(-4) m. Using the formula for the cross-sectional area of a wire, A = πr², where r is the radius, we find that A = π x (4.1 x 10^(-4) m / 2)².

Now, we can find the current that the wire can carry using the formula I = J x A, where J is the current density. Rearranging the formula to solve for J, we get J = I / A. Plugging in the given values, we have J = 1 A / (π x (4.1 x 10^(-4) m / 2)²).

By comparing the calculated currents for both wires, you can determine which wire to use.