A cylindrical wire of radius 2.0 mm carries a current of 2.5 A. The potential 4) difference between points on the wire that are 46 m apart is 3.7 V.
The correct answer and explanation is:
To determine the resistivity of the cylindrical wire, we will use Ohm’s Law and the formula for the resistance of a cylindrical conductor.
Correct Answer:
The resistivity of the wire is approximately 4.03 × 10⁻⁷ Ω·m
Explanation (300 words):
Electric current flowing through a conductor encounters resistance. This resistance causes a voltage drop along the length of the conductor. In this problem, a cylindrical wire has a known current, length, radius, and voltage drop. Using these values, the resistivity of the material can be calculated.
First, recall Ohm’s Law: V=IRV = IR
Where:
- VV is the potential difference (3.7 V),
- II is the current (2.5 A),
- RR is the resistance over the 46 m section.
Rearranging the equation: R=VI=3.72.5=1.48 ΩR = \frac{V}{I} = \frac{3.7}{2.5} = 1.48\ \Omega
Next, use the formula for resistance of a cylindrical conductor: R=ρLAR = \frac{\rho L}{A}
Where:
- ρ\rho is the resistivity,
- LL is the length (46 m),
- AA is the cross-sectional area of the wire,
- A=πr2=π(0.002)2=1.2566×10−5 m2A = \pi r^2 = \pi (0.002)^2 = 1.2566 \times 10^{-5}\ m^2
Now, rearranging to solve for resistivity: ρ=RAL\rho = \frac{RA}{L} ρ=1.48×1.2566×10−546\rho = \frac{1.48 \times 1.2566 \times 10^{-5}}{46} ρ≈1.860×10−546≈4.04×10−7 Ω⋅m\rho \approx \frac{1.860 \times 10^{-5}}{46} \approx 4.04 \times 10^{-7}\ \Omega\cdot m
So, the resistivity of the wire material is approximately 4.04 × 10⁻⁷ ohm-meters.
This value is consistent with the resistivity of materials like aluminum, indicating the wire might be made from that. Knowing resistivity helps identify materials and evaluate their effectiveness in conducting electricity. Lower resistivity means the material is a better conductor.