Question

For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0400 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 13.0 cm long. Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)

Answer 1

Answer:

n = 16

Explanation:

given,

magnetic field = 0.04 T

current = 1 A

diameter of wire = 0.5 mm

length of solenoid = 13 cm = 0.13 m

number of layers = ?

we know,

Magnetic field = permeability x turn density x current

$B = \dfrac{\mu_0IN}{L_{cylinder}}\times n$

where n is the number  of layer

N = L/d

$B = \dfrac{\mu_0 I \dfrac{L_{cylinder}}{d_{wire}}}{L_{cylinder}}\times n$

$B = \dfrac{\mu_0 I}{d_{wire}}\times n$

$n = \dfrac{d_{wire}\ B}{\mu_0 I}$

$n = \dfrac{0.5\times 10^{-3}\times 0.04}{4\pi \times 10^{-7}\times 1}$

n = 15.91 = 16 (approx)

number of layers is equal to 16

Answer 2

The number of layers of the wire needed for the solenoid project is 16 layers.

The given parameters;

• magnitude of the magnetic field, B = 0.04 T
• current in the solenoid, I = 1 A
• diameter of the wire, d = 0.5 mm = 0.5 x 10⁻³ m
• length of the cylinder, L = 13 cm = 0.13 m

The formula for calculating magnetic field strength (B) for a given number of  layers (n) of solenoid wire is given as;

$B = \frac{\mu_o I n}{d_{wire}} \\\\n = \frac{B \times d_{wire}}{\mu_0 I}$

where;

n is the number of layers of wire needed;

Substitute the given values and solve for the number of layers of wire needed.

$n = \frac{0.04\times 0.5\times 10^{-3}}{(4\pi \times 10^{-7}) \times (1)} \\\\n = 15.91 \\\\n\approx 16 \ layers$

Thus, the number of layers of the wire needed for the solenoid project is 16 layers.

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