A copper wire has diameter \(0.5 \mathrm{~mm}\) and resistivity of \(1.6 \times 10^{-8} \Omega \mathrm{m}\). What will be the length of this wire to make its resistance \(10 \Omega ?\) How much does the resistance change if the diameter is doubled?

Ohm's Law is a fundamental principle in electronics and electrical engineering. It helps us understand how voltage, resistance, and current are related in an electrical circuit. The law is written as:
\( V = IR \)
where:

• \( V \) is the voltage (in volts),
• \( I \) is the current (in amperes),
• \( R \) is the resistance (in ohms).

In this exercise, we are more interested in a rearranged form of Ohm's Law, which allows us to find the resistance:
\( R = \frac{V}{I} \).
However, when working with materials and their specific properties, such as a copper wire in this case, we use a modified version:
\( R = \rho \frac{L}{A} \),
where:

• \( \rho \) is the material's resistivity (in ohm-meters),
• \( L \) is the length of the wire (in meters),
• \( A \) is the cross-sectional area of the wire (in square meters).

Resistivity is a material-specific property that describes how strongly a material opposes the flow of electric current. It's denoted by \( \rho \) and measured in ohm-meters (\( \Omega \cdot m \)).
A lower resistivity means that the material allows electric current to flow more easily. Copper, for example, has a very low resistivity of \( 1.6 \times 10^{-8} \Omega \cdot m \), which makes it an excellent conductor.
To find the resistance of a piece of wire made from a specific material, we use the formula:
\( R = \rho \frac{L}{A} \).
This emphasizes that both the material’s resistivity and the dimensions of the wire (length \( L \) and cross-sectional area \( A \)) affect the total resistance.

To find the resistance of a wire, we need to know its cross-sectional area \( A \). For a wire with a circular cross-section (which is most common in electrical wires), the cross-sectional area can be calculated using the formula:
\( A = \pi r^2 \),
where \( r \) is the radius of the wire.
Given the diameter \( d \) of the wire, the radius \( r \) is half of the diameter:
\( r = \frac{d}{2} \).
In this exercise, the diameter is 0.5 mm, so:
\( r = \frac{0.5 \times 10^{-3} m}{2} = 0.25 \times 10^{-3} m \).
Substituting \( r \) into the area formula:
\( A = \pi (0.25 \times 10^{-3} m)^2 = 1.9635 \times 10^{-7} m^2 \).
This cross-sectional area is then used to calculate the length of the wire needed to achieve a desired resistance.

The diameter of a wire significantly affects its resistance. A smaller diameter results in a higher resistance, while a larger diameter results in a lower resistance. This relationship is because the cross-sectional area \( A \) is directly proportional to the square of the radius \( r \):
\( A = \pi r^2 \).
If the diameter of the wire is doubled, the radius is also doubled, which means the new cross-sectional area is:
\( A_{new} = \pi (2r)^2 = 4\pi r^2 \).
This shows that doubling the diameter quadruples the cross-sectional area.
Therefore, with the same length \( L \), the new resistance \( R_{new} \) is one-fourth of the original resistance. In our example, with a starting resistance of 10 ohms, after doubling the diameter, the new resistance is approximately 2.5 ohms, demonstrating this inverse relationship.