Your employer asks you to build a 17-Cm-long solenoid with an interior field of

Question

Your employer asks you to build a 17-Cm-long solenoid with an interior field of 4.3mT. 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 #18 gauge has a diameter of 1.02mm and has a maximum current rating of 6 A. Wire with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A. Part A Which wire should you use? #18 #26 Correct Part B What current will you need? Express your answer to two significant figures and include the appropriate units. I = Value Units

Explanation / Answer

As we know that ,

The magnetic field due to the solenoid

B = ?0NI / L .................(1)

Here N = number of turns

L = length of the solenoid = 0.17 m

magnetic field B = 4.3 x10^-3 T

?0   =   4? x10^-7 T   A/m

If the wires are wound as closely as possible ,

the spacing between one turn and the next is simply

the diameter of the wire d = 1.02 mm = 1.02x10-3 m

The number of tuns that can fit into the length L is

For (#18 guage ) wire

N = L / d =   0.17m / ( 1.02 x 10^-3 m)  

    = 166.67

rearraging the equation (1) is

we have ,

I =   BL / ?0N

plug all values we get

   I =    ( 4.3 x10^-3)(0.17) / (4? x10^-7)(166.67)

       = 3.49 A

futher ,

  For   (#26 gauge wire )

Number of turns    N =   L /d    = 0.17 m / (0.41 *10^-3m )

                                    = 414.6 turns

                                   

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