Your employer asks you to build a 31-cm-long solenoid with an interior field of

Question

Your employer asks you to build a 31-cm-long solenoid with an interior field of 5.2mT . 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 1.02 mm 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. Which wire should you use? #18 #26 Part B What current will you need? Express your answer to two significant figures and include the appropriate units. I

Explanation / Answer

Magnetic field due to a solenoid

B=uo*N*I/L

=>I=B*L/uo*N

For 18 Gauge Wire

N=L/d= 0.31/(1.02*10-3) =304 turns

I=(5.2*10-3)*(0.31)/(4pi*10-7)*304

I=4.2 A

For 26 gauge wire

N=0.31/(0.41*10-3) =757 turns

I=(5.2*10-3)*(0.31)/(4pi*10-7)*757

I=1.7 A

a)

For 26 gauge wire I=1.69 A>1 A Limit of the wire .so 18 gauge wire must be used.

b)

I=4.2 A

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