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------------------------------------------------ A solenoid 1.55 m long and 2.80 cm in diameter carries a current of 21.0 A. The magnetic field inside the solenoid is 17.0 mT. Find the length of the wire forming the solenoid.

To find the length of the wire forming the solenoid, we need to determine the number of turns (N) in the solenoid first. We can use the formula for the magnetic field inside a solenoid:

B = μ₀ * (N / L) * I

Where B is the magnetic field (17.0 mT or 0.017 T), μ₀ is the permeability of free space (4π x [tex]10^{-7}[/tex] Tm/A), L is the length of the solenoid (1.55 m), and I is the current (21.0 A). Rearranging the formula for N:

N = (B * L) / (μ₀ * I)

Now, plug in the given values:

N ≈ (0.017 T * 1.55 m) / (4π x [tex]10^{-7}[/tex] Tm/A * 21.0 A)
N ≈ 409.5

Since the number of turns must be a whole number, we can round it to the nearest whole number, N ≈ 410 turns.

Next, we'll find the length of the wire. The circumference of the solenoid is given by:

C = π * D

Where D is the diameter (2.80 cm or 0.028 m). Calculating the circumference:

C ≈ π * 0.028 m
C ≈ 0.088 m

Finally, we multiply the circumference by the number of turns to get the total length of the wire:

Length ≈ 410 turns * 0.088 m/turn
Length ≈ 36.08 m

So, the length of the wire forming the solenoid is approximately 36.08 meters.

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