A coil is formed by winding 190 turns of insulated 16 gauge copper wire (diameter = 1.9 mm) in a single layer on a cylindrical form of radius 9.2 cm. What is the resistance of the coil? Neglect the thickness of the insulation. (Take the resistivity of copper to be 1.69 × 10-8 ohm-m.)

Explanation:

The given data is as follows.

number of turns = n = 190 turns

radius = r = 9.2 cm = 0.092 m

diameter of copper wire = d = 1.9 mm

radius of copper wire = =

                     =    

                     = m

where,  A = cross sectional area of the wire

As, length of each turn of wire is 2pr where r is radius of the coil.

  L = 190(2pr)

     =

      = 109.77 m

The cross sectional area of the wire is as follows.

       A =

           =

          =

It is given that resistivity of copper wire is

              R =

                  =

                 =

Thus, we can conclude that resistance of the coil is .

The resistance of the coil is 0.655 Ω

Explanation:

Given that,

Number of turns = 190 turns

Diameter of coil= 1.9 mm = 0.95 mm

Radius of single layer = 9.2 cm

Pressure = 16 gauge

We need to calculate the length of the wire

Using formula of length

Where, n = number of turns

r = radius of cylinder

Put the value into the formula

We need to calculate the area of cross section

Using formula of area

Put the value into the formula

We need to calculate the resistance of the coil

Using formula of resistivity

Put the value into the formula

Hence, The resistance of the coil is 0.655 Ω

1/0.14nm = 10^10/1.4 a lot