Question.
A copper wire has a diameter of $0.5 \mathrm{~mm}$ and a resistivity of $1.6 \times 10^{-6} \mathrm{ohm} \mathrm{cm}$. How much of this wire would be required to make a 10 ohm coil? How much does the resistance change if the diameter is doubled?
A copper wire has a diameter of $0.5 \mathrm{~mm}$ and a resistivity of $1.6 \times 10^{-6} \mathrm{ohm} \mathrm{cm}$. How much of this wire would be required to make a 10 ohm coil? How much does the resistance change if the diameter is doubled?
solution:
We are given that, diameter of the wire,
$D=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}$
Radius $\mathrm{r}=\frac{\mathrm{D}}{2}=\frac{0.5 \times 10^{-3}}{2}=2.5 \times 10^{-4} \mathrm{~m}$
Resistivity of copper, $\rho=1.6 \times 10^{-6} \mathrm{ohm} \mathrm{cm}$
$=1.6 \times 10^{-8}$ ohm $\mathrm{m}$
Required resistance, $\mathrm{R}=10$ ohm
As, $\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}, \ell=\frac{\mathrm{RA}}{\rho}=\frac{\mathrm{R}\left(\pi \mathrm{r}^{2}\right)}{\rho} \quad\left[\mathrm{A}=\pi \mathrm{r}^{2}\right]$
or $\ell=\frac{3.14 \times 10 \times\left(2.5 \times 10^{-4}\right)^{2}}{1.6 \times 10^{-8}}=112.7 \mathrm{~m}$
Since, $\quad \mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \ell}{\pi \mathrm{r}^{2}}=\frac{\rho \ell}{\pi(\mathrm{D} / 2)^{2}}=\frac{4 \rho \ell}{\pi \mathrm{D}^{2}}, \mathrm{R} \propto \frac{1}{\mathrm{D}^{2}}$
When D is doubled, R becomes $\frac{1}{4}$ times.
We are given that, diameter of the wire,
$D=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}$
Radius $\mathrm{r}=\frac{\mathrm{D}}{2}=\frac{0.5 \times 10^{-3}}{2}=2.5 \times 10^{-4} \mathrm{~m}$
Resistivity of copper, $\rho=1.6 \times 10^{-6} \mathrm{ohm} \mathrm{cm}$
$=1.6 \times 10^{-8}$ ohm $\mathrm{m}$
Required resistance, $\mathrm{R}=10$ ohm
As, $\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}, \ell=\frac{\mathrm{RA}}{\rho}=\frac{\mathrm{R}\left(\pi \mathrm{r}^{2}\right)}{\rho} \quad\left[\mathrm{A}=\pi \mathrm{r}^{2}\right]$
or $\ell=\frac{3.14 \times 10 \times\left(2.5 \times 10^{-4}\right)^{2}}{1.6 \times 10^{-8}}=112.7 \mathrm{~m}$
Since, $\quad \mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \ell}{\pi \mathrm{r}^{2}}=\frac{\rho \ell}{\pi(\mathrm{D} / 2)^{2}}=\frac{4 \rho \ell}{\pi \mathrm{D}^{2}}, \mathrm{R} \propto \frac{1}{\mathrm{D}^{2}}$
When D is doubled, R becomes $\frac{1}{4}$ times.