-1 points SerCP10 19.P061 My Notes Ask Your Teac t is desired to construct a sol

Question

-1 points SerCP10 19.P061 My Notes Ask Your Teac t is desired to construct a solenoid that will have a resistance of 5.60 (at 20°C) and produce a magnetic field of 4.00 × 10-2 T at its center when it carries a current of 3.15 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°C is 1.7 × 1 0-8 ·m) (a) the number of turns of wire needed to build the solenoid turns (b) the length the solenoid should have cm

Explanation / Answer

we know that the magnetic field (B) inside of a solenoid is given by the equation,
B = µo* (N / L) * I, n=N/L

also the resistance (R) of a piece of wire with cross sectional area (A) is given by,
R = ( * l) / A
Where is the resistivity of the wire,

l is the total length, and

A is the cross sectional area

  Now with the given resistance R = 5.60

R = *L/A

So L = R*A/ = 5.60**(0.25x10-3)2/(1.7x10-8 )= 64.6 m

however each turn of the solenoid is d = *2*0.01m = 6.28x10-2 m

so there are 64.6/6.28x10-2 = 1028 turns

b) B= µo n*i
so n = B/(µo*i) = 4x10-2 (4x10-7*3.15) = 10110 turns per unit length

we now have the number of turn =1028 and the number of turns per unit length is 10110

so the length of the solenoid must be 1028/10110 = 0.102 m = 10.2 cm

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