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You want to wind a solenoid that is 3.5 cm in diameter, is 23 cm long, and will have a magnetic field of 0.0250 T when a current of 3.0 A is in the wire that makes up the solenoid. 1) What total length of wire do you need?

Respuesta :

Answer:

L'=167.7 m

Explanation:

Given that

d= 3.5 cm

L= 23 cm

B=0.025 T

I= 3 A

As we know that magnetic field in solenoid given as

[tex]B=\dfrac{\mu _oNI}{L}[/tex]

N=Number of turns

L=Length

B=Magnetic filed

I=Current

[tex]N=\dfrac{BL}{\mu _oI}[/tex]

Now by putting the all values

[tex]N=\dfrac{BL}{\mu _oI}[/tex]

[tex]N=\dfrac{0.025\times 0.23}{4\pi \times 10^{-7}\times 3}[/tex]

N=1525.23 turns = 1526 turns

Total length L'=Nπd

L'=1526 x 3.14 x 0.035

L'=167.7 m