Question

A wire has a diameter of $0.5mm$ and a resistivity of $1.6\times {10}^{-6}\Omega cm$. How much of this wire would be required to make a $10\Omega$ coil?

• A. $1000m$
• B. $57m$
• C. $157.6m$
• D. $122.7m$

Answer 1

The correct option is D $122.7m$

Given:
Resistivity, $\rho =1.6\times {10}^{-6}\Omega cm$
Converting to S.I. units, $\rho =1.6\times {10}^{-8}\Omega m$
Diameter of wire, $D=0.5mm$
Resistance of wire, $R=10\Omega$

Cross section area of wire will be:
$A=\pi r^2$
$A=\pi {\left(\frac{D}{2}\right)}^2$
$A=\frac{\pi }{4}\times (0.5\times {10}^{-3}m{)}^2$
$A=\frac{\pi }{16}\times {10}^{-6}{m}^2$

Now, resistance is given by:
$R=\rho \frac{L}{A}$
$L=\frac{RA}{\rho }$
$L=\frac{10\times \frac{\pi }{16}\times {10}^{-6}}{1.6\times {10}^{-8}}$
$L\approx 122.7m$