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A ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

Short Answer

Expert verified
The resistance of the wire when stretched to three times its original length, assuming the density and resistivity remain constant, is \(9 \times R\).

Step by step solution

01

Recognize the formula for resistance

The resistance \(R\) of a wire can be calculated using the formula \(R=\rho * \frac{l}{A}\) where: \( \rho\) is the resistivity of the material, \( l \) is the length and \( A \) is the cross-sectional area of the wire.
02

Understand the change in wire dimension

When the wire is stretched to three times its original length, its volume remains constant. That is, the original length (\(l\)) times the original cross-sectional area (\(A\)) is equal to the new length (3\(l\)) times the new cross-sectional area (\(A'\)) . This gives us: \( l*A = 3*l*A'\). Solving this for \(A'\) we find that \(A' = \frac{A}{3}\).
03

Calculate the New Resistance

The new resistance \(R'\) is obtained by substituting for \(l\) and \(A'\) in the formula obtained in step 1, giving \(R' = \rho * \frac{3l}{\frac{A}{3}} = R * 9\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistivity
Resistivity is an intrinsic property of a material that measures its opposition to the flow of electric current. It is denoted by the symbol \(\rho\). Imagine resistivity as the inherent friction an electric charge feels as it travels through a wire.
This property is not affected by the size or shape of the material itself. It depends solely on the type of material from which the wire is made. Metals like copper and aluminum have low resistivity, which makes them excellent conductors.
  • Resistivity is measured in ohm-meters (\(\Omega \cdot m\)).
  • It is affected by temperature, usually increasing as the temperature rises.
In our exercise, when the wire is stretched, the resistivity remains unchanged, which simplifies calculations.
Cross-sectional area
The cross-sectional area of a wire represents the size of its face when cut perpendicular to its length. Think of it as the thickness of the wire. This area plays a crucial role because it determines how much space is available for electrical current to pass through.
In the provided exercise, when the ductile wire is stretched to three times its original length, the cross-sectional area is significantly reduced. This is because the volume of the wire remains constant — stretching increases length while decreasing the cross-sectional area.Understanding this, we find that if the length is tripled, then the new cross-sectional area \(A'\) is reduced to a third of the original (\(A' = \frac{A}{3}\)). Consequently, the smaller the cross-sectional area, the higher the resistance.
Ductile metal wire
A ductile metal wire is one that can be easily stretched or drawn into thin threads without breaking. Metals like copper, aluminum, and gold are examples that exhibit high ductility, making them perfect for wiring and cables.
  • Ductility involves the ability to withstand tensile stress.
  • It is a property that differs from resistivity, focusing on the material's physical flexibility instead.
In the context of the exercise, the fact that the wire can be stretched to three times its original length means it must be quite ductile. When stretched, the physical dimensions change, but the amount of material remains constant. This affects the geometry but not the material's inherent properties, such as resistivity.
Electrical resistance calculation
Electrical resistance is a measure of how much a conductor opposes the flow of electric current. The formula for resistance \(R\) is \(R = \rho \cdot \frac{l}{A}\), where \(\rho\) is the resistivity, \(l\) is the length of the conductor, and \(A\) is the cross-sectional area.
To calculate the new resistance when a wire is stretched, we must adjust the values of \(l\) and \(A\) accordingly. In our exercise:
  • Original length is \(l\); new length is \(3l\).
  • Original area is \(A\); new area is \(\frac{A}{3}\).
Substituting these into the resistance formula gives us the new resistance \(R'\), which is \(R' = \rho \cdot \frac{3l}{\frac{A}{3}} = 9R\). Hence, stretching the wire increases its resistance ninefold.

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Most popular questions from this chapter

A circular loop of wire with radius \(2.00 \mathrm{~cm}\) and resistance \(0.600 \Omega\) is in a region of a spatially uniform magnetic field \(\vec{B}\) that is perpendicular to the plane of the loop. At \(t=0\) the magnetic field has magnitude \(B_{0}=3.00 \mathrm{~T}\). The magnetic field then decreases according to the equation \(B(t)=B_{0} e^{-t / \tau},\) where \(\tau=0.500 \mathrm{~s}\). (a) What is the maximum magnitude of the current \(I\) induced in the loop? (b) What is the induced current \(I\) when \(t=1.50 \mathrm{~s} ?\)

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