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A 20.0-m-long copper wire, 2.00 mm in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid of diameter 2.50 cm (outer edge). What is

(a) the length of the solenoid and
(b) the field at the center when the current in the wire is 16.7 A

Respuesta :

To solve this problem it is necessary to apply the concepts related to the magnetic field within a solenoid.

The magnetic field from the Ampere law is defined as

[tex]B = \mu n I[/tex]

Where

[tex]\mu_0[/tex]= Permeability constant

I = Current

n = number of loops per length unit

At the same time the number of loops would be subject to the length of the solenoid by the number of turns it takes on it.

[tex]N = \frac{L}{2\pi r}[/tex]

Where,

L = Length

r = Radius

Part A) The total number of laps would be

[tex]N = \frac{L}{2\pi r}[/tex]

[tex]N = \frac{20}{2\pi (1.25*10^{-2})}[/tex]

[tex]N = 255Turn[/tex]

Therefore the effective length would be

[tex]l = N*d[/tex]

[tex]l = 255*(2*10^{-3})[/tex]

[tex]l = 0.509m[/tex]

Therefore the length of the solenoid is 0.509m

Part B) The value of the magnetic field would be given from the first evaluation for which

[tex]B = \mu n I[/tex]

[tex]B = (4\pi*10^{-7}) (\frac{255}{0.509})(16.7)[/tex]

[tex]B = 10.5mT[/tex]

Therefore the field at the center when the current in the wire is 16.7 is 10.5mT