It is desired to construct a solenoid that will have a resistance of 5.80 ohm (a

Question

It is desired to construct a solenoid that will have a resistance of 5.80 ohm (at 20 degree C) and produce a magnetic field of 4.00 times 10^-2 T at its center when it carries a current of 3.20 A. The solenoid is to be constructed from copper wire having a diameter of 0.500 mm. If the radius of the solenoid is to be 1.00 cm, determine the following. (The resistivity of copper at 20 degree C is 1.7 times 10^-8 ohm middot m.) (a) the number of turns of wire needed to build the solenoid turns (b) the length the solenoid should have cm

Explanation / Answer

a) Now with the R = 5.80 we have R = *L/A

So L = R*A/ = 5.80**(0.25x10^-3m)^2/1.70x10^-8 = 66.989m

Each turn of the solenoid is d = *2*0.01m = 6.28x10^-2m

so there are 66.989/6.28x10^-2 = 1066.70= 1067 turns

b) The magnetic field B = µo*n*i where n = turns per unit length

so n = B/(µo*i) = 4.00x10^-2/(4x10^-7*3.200) = 9947 turns per unit length

We have the number of turn = 1067 and the necessary turns per unit length is 9947

so the length of the solenoid must be 1067/9947 = 0.10726m = 10.726cm

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