Question

A silver wire has a diameter of 0.4 mm and resistivity 1.6×10^-8 ohm metre. How much length of the wire is required to make a 1 ohm coil?

Answer 1

Resistance= 1ohm
RESISTIVITY=1.6 * 10-8
DIAMETER=0.4mm
. . . Radius=0.4/2=0.2mm = 0.2/1000 m = 0.2 * 10-3
         (R)2=(0.2 * 10-3)2
        R2=0.04 * 10-6
Area= pie * R2=22/7  * 0.04*10-6  
                      = 1.25714286 * 10-6
                       = 1.26 * 10-6 m2
NOW Resistance = resistivity * length / area
              LENGTH= RESISTANCE * AREA / RESISTIVITY
            LENGTH=1 * 1.26 * 10-6 /1.6 * 10-8
                           = 126 * 108 *10/16 * 100 * 106
                          =1260/16
           LENGTH = 78.75 m

Answer 2

Here we are given with :

Diameter of silver wire , D = 0.4 mm

Radius of wire = D/2 = 0.4 mm / 2 = 0.2 mm

= 0.2 × 10^-3 m

Resistivity = p = 1.6 × 10^-8 Ω-m

Resistance , R = 1 ohm = 1 Ω

Length is to be found .

Using the formula :

$R = p \frac{l}{A } = p \frac{l}{\pi {r}^{2} } \\ \\ l = \frac{\pi \: R\: {r}^{2} }{p} = \frac{22}{7} \times \frac{1 \times {(0.2 \times {10}^{ - 3} )}^{2} }{1.6 \times {10}^{8} } = 7.8 \: m$

The required length is 7.86 m .