Question

A 17.0 m -long copper wire, 3.00 mm in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid of diameter 2.80 cm (outer edge).

What is the length of the solenoid?

What is the field at the center when the current in the wire is 17.0 A ? show work

Answer #1

Outer edge diameter of solenoid = 28 mm

diameter of wire = 3mm

diameter of loop formed by center of wire d = 25mm

let length of one turn of loop be ' l ' . l = pi d

total number of turns N = total length of wire/ length of one turn
= L/l

Length of solenoid = Number of turns x thickness of one turn

thickness of one turn is diameter of wire = 3mm

hence length of solenoid = ( L/ l) 3mm = (17000/pi x 25 ) x3 =
649.7 mm = 64.97 cm

Magnetic field at center = (mu not) n i (mu not is permeability
of vacuum, n is no. of turns per unit length and i is current

n = 1/ diameter of wire = 1000/3

B = (4 pi x 10^{-7}) 17 x 1000 /3 = 7.1 x 10^{-5}
T

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