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Q15P

Expert-verifiedFound in: Page 766

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A coil is formed by winding 250turns**** of insulated 16-gauge copper wire (diameter = 1.3 mm****) in a single layer on a cylindrical form of radius 12cm****. What is the resistance of the coil? Neglect the thickness of the insulation. (Use Table 26-1.)**

The resistance of the coil is 2.4$\mathrm{\Omega}$.

- Number of turns of coil,
*n*= 250 - Diameter of the wire, $d=1.3\mathrm{mm}\mathrm{or}1.3\times {10}^{-3}\mathrm{m}$
- Radius of coil, $R=12\mathrm{cm}\mathrm{or}12\times {10}^{-2}\mathrm{m}$

**The voltage is directly proportional to the current and the constant of proportionality is called Resistance. This is called Ohm’s law.**

**We can find the total length of the coil by using the radius of the coil. We can also find the area of the cross-section of the wire from its diameter. Then, by using these values in the formula for resistance, we can find the resistance of the copper coil.**

Formulae:

The resistance of a wire due, $R=\frac{pL}{A}$ …(i)

Here, *R* is resistance, p is resistivity, *L* is length, *A *is the area of cross-section.

The circumference of the circle, $L=2\mathrm{\pi}R$ …(ii)

is circumference, is the radius of the circle.

The cross-sectional area of a circle, $A={\mathrm{\pi R}}^{2}$ …(iii)

Length of one turn of the coil can be given using the circumference of the circle.

So, the length of 250 turns of the coil using equation (ii) would be,

$L=250\times \left(2\mathrm{\pi R}\right)\phantom{\rule{0ex}{0ex}}=250\times 2(3.142)(12\times {10}^{-2}\mathrm{m})\phantom{\rule{0ex}{0ex}}=188.49\mathrm{m}$

Similarly, the cross sectional area of the wire is given using equation (iii) as follows:

$A=\mathrm{\pi}\frac{{\mathrm{d}}^{2}}{4}(\therefore \mathrm{r}=\mathrm{d}/2)\phantom{\rule{0ex}{0ex}}=(3.142)\frac{{(1.3\times {10}^{-3}\mathrm{m})}^{2}}{4}\phantom{\rule{0ex}{0ex}}=1.327\times {10}^{-6}{\mathrm{m}}^{2}$

Resistivity of copper is $\mathrm{p}=1.69\times {10}^{-8}\mathrm{\Omega m}$

Now, using these above values in equation (i); the resistance of the copper coil can be calculated as follows:

$\mathrm{R}=\frac{(1.69\times {10}^{-8}\mathrm{\Omega m})(188.49\mathrm{m})}{1.327\times {10}^{-6}{\mathrm{m}}^{2}}\phantom{\rule{0ex}{0ex}}=2.4\mathrm{\Omega}$

Therefore, the resistance of the copper coil is $2.4\mathrm{\Omega}$.

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