(b)i. Mention and state the law on which the experiment in (a) is based.
ii. A piece of resistance vire of diameter 0.2 mm and resistance m has a resistivity of 8.8 x 10\(^{-7}\)\(\Omega\)m, calculate the length of the Wire. [\(\pi\) =\(\frac{22}{7}\)]
Table of values
XY = 100cm
| R(\(\Omega\)) | 1(amp) | V |
| 0 | 0.15 | 2.65 |
| 1 | 0.15 | 2.65 |
| 5 | 0.15 | 2.65 |
| 10 | 0.10 | 1.85 |
| 20 | 0.08 | 1.40 |
| 40 | 0.06 | 0.90 |
| 60 | 0.04 | 0.70 |
Slope = \(\frac{x_{2}-x_{1}}{y_{2}-y_{1}} = \frac{1.75-0.5}{0.10-0.025} = \frac{1.25}{0.075}\)
Precautions:
(b)i. The experiment is based on Ohm's law. Ohm's law states that the current flowing through a metallic conductor is proportional to the potential difference between its ends provided the temperature and other physical conditions of the conductor remain constant; i.e: \(\frac{V}{|}\) = Constant.
ii. Area of cross-section of wire = \(\frac{\pi d^{2}}{4} = \frac{\frac{22}{7} \times (0.2)^{2}}{4}\)
\(\frac{3.143 \times 0.04}{4}\) = 0.031mm\(^{2}\) = 3.1m\(^{2}\)
Resistivity = \(\frac{RA}{L}\), length (L) = \(\frac{R.A}{resistivity}\)
= \(\frac{7 \times 3.1}{8.8 \times 10^{-7}} = \frac{21.7}{8.8 \times 10^{-7}}\) = 2.47 x 10\(^{-7}\)
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