Question

Third Test, Page 4 of 6 Nov 7, 2016 3. To construct a solenoid, you wrap insulatedl wire umiformly around a plastic tube 7 6em in diameter and 33em in length. You would like a 3.0A current to produce a 2.5kG magnetic field inside your solenoid. What is the total length of wire you will ned to meet these specifications? (1) (5 points) To find the length of wire, first find the density of loops per unit of length in your solenoid. (II) (5 points) Find the length of a single loop of wire. (III) (10 points) Using the previous two results, find the number of loops in your solenoid and then the total length of wire needed to meet the specifications.

Answer 1

3)

given

L = 7.6 cm = 0.076 m
d = 33 cm = 0.33 m
I = 3 A
B = 2.5 kG = 2.5*10^3*10^-4 = 0.25 T

a) let n is the number of turns per unit length.

we know, B = mue*n*I

==> n = B/(mue*I)

= 0.25/(4*pi*10^-7*3)

= 66314 turns/m

b) length of single loop of wire = 2*pi*r

= 2*pi*(d/2)

= 2*pi*(0.33/2)

= 1.037 m

c) no of turns in the given in the solenoid, N = n*L

= 66314*0.076

= 5040 turns

length of the wire needed = N*1.037

= 5040*1.037

= 5226 m