An electric heating element to dissipate 480 watts on 240 V mains is to be made from nichrome wire of 1 mm diameter. Calculate the length of the wire required if the resistivity of nichrome is 1.1 × 10-6 Ω m.

An electric heating element to dissipate 480 watts on 240 V mains is to be made from nichrome wire of 1 mm diameter. Calculate the length of the wire required if the resistivity of nichrome is 1.1 $\times$ 10${}^{-6}$ $\mathrm{\Omega }m$.
Solution:
Given,
power dissipation, $P$ = 480 W
voltage supply, $V$ = 240 V
diameter of wire, $d$ = 1 mm radius of wire, $r$ = 0.5 mm = 0.5 $\times$ 10${}^{-3}$ m
length of wire, $l$=?
resistivity of wire, $\rho$ = 1.1 $\times$ 10${}^{-6}$ $\mathrm{\Omega }m$

Now, $$\begin{array}{rl}P& =\frac{V^2}{R}\\ R& =\frac{V^2}{P}\\ & =\frac{(240{)}^2}{480}\\ \therefore R& =120\mathrm{\Omega }\end{array}$$ Again,
$$\begin{array}{rl}R& =\frac{\rho l}{A}\\ l& =\frac{RA}{\rho }\\ & =\frac{120\times \pi \times (0.5\times {10}^{-3}{)}^2}{1.1\times {10}^{-6}}\\ \therefore l& =85.68m\end{array}$$ Hence, the required length of wire is 85.68 m.