Question

A copper wire has diameter 0.5 mm and resistivity of 1.6*10power - 6cm how much of this wire would be required to make a 10ohm coil

Answer 1

Given:

Diameter = 0.5 mm = 0.5 ×$10^{-3}$

Resistivity = 1.6 × $10^{-6}$ Siemens per centimeter

= 1.6 × $10^{-4}$ Siemens per meter

Resistance = 10 Ohm

To find:

Length of wire = ?

Formula used:

Resistance of the wire is given by,

R = ω $\frac{l}{A}$

Where R is the resistance

ω = resistivity

A = area of cross section

l = length of wire

Solution:

Area of cross section of a wire = Area of the circle

A = π$r^{2}$

A = 3.14 × $\frac{0.5}{2} \times10^{-3} \times \frac{0.5}{2} \times 10^{-3}$

A = 1.9625 × $10^{-7}$

Resistance of the wire is given by,

R = ω $\frac{l}{A}$

Where R is the resistance

ω = resistivity

A = area of cross section

l = length of wire

l = $\frac{RA}{ω}$

l = $\frac {1.9625 \times 10^{-7} \times 10}{1.6 \times 10^{-4} }$

l = 12.26 ×$10^{-3}$ meter

l = 12.26 millimeter

Thus, length of wire required to make 10 Ohm coil is 12.26 mm.