Problem 80 - University Physics with Modern Physics

Question

A lightning bolt strikes one end of a steel lightning rod, producing a \(15,000-\) A current burst that lasts for 65 \mus. The rod is 2.0 \(\mathrm{m}\) long and 1.8 \(\mathrm{cm}\) in diameter, and its other end is connected to the ground by 35 \(\mathrm{m}\) of \(8.0-\mathrm{mm}\) -diameter copper wire. (a) Find the potential difference between the top of the steel rod and the lower end of the copper wire during the current burst. (b) Find the total energy deposited in the rod and wire by the current burst.

Step-by-Step Solution

Answer

(a) The potential difference is approximately 6.81 kV. (b) The total energy deposited is around 990 J.

Step 1: Calculate the resistance of the steel rod

First, calculate the resistance of the steel rod using the formula for resistance: \[ R = \frac{\rho L}{A} \] where \( \rho_{\text{steel}} = 1.68 \times 10^{-7} \, \Omega \cdot \text{m} \) is the resistivity of steel, \( L = 2.0 \, \text{m} \) is the length, and \( A \) is the cross-sectional area. The area can be calculated as \( A = \pi \left(\frac{d}{2}\right)^2 \), with \( d = 0.018 \, \text{m} \) being the diameter.

Step 2: Calculate the resistance of the copper wire

Similarly, calculate the resistance of the copper wire:\[ R = \frac{\rho L}{A} \] where \( \rho_{\text{copper}} = 1.72 \times 10^{-8} \, \Omega \cdot \text{m} \), \( L = 35 \, \text{m} \) is the length, and \( A \) is the cross-sectional area. The area for copper is \( A = \pi \left(\frac{d}{2}\right)^2 \), with \( d = 0.008 \, \text{m} \).

Step 3: Calculate the total resistance

Add the resistances of the steel rod and copper wire to find the total resistance:\[ R_{\text{total}} = R_{\text{steel}} + R_{\text{copper}} \]

Step 4: Determine the potential difference

Use Ohm's Law to find the potential difference \( V \):\[ V = I \times R_{\text{total}} \]where \( I = 15000 \, \text{A} \) is the current.

Step 5: Calculate the total energy deposited

Use the formula for energy \( E \) to find the total energy deposited:\[ E = I^2 \times R_{\text{total}} \times t \]where \( t = 65 \times 10^{-6} \, \text{s} \) is the duration of the current.

Key Concepts

Resistance Calculation

Resistance is a fundamental concept in electrical circuits that characterizes how much a material opposes the flow of electric current. To calculate the resistance, we use the formula: \[ R = \frac{\rho L}{A} \] where:

• \( R \) is resistance,
• \( \rho \) is resistivity (a unique property of each material),
• \( L \) is the length of the material,
• \( A \) is the cross-sectional area.

In our example, we calculate the resistance of both the steel rod and the copper wire:

• For the steel rod with resistivity \( \rho_{\text{steel}} = 1.68 \times 10^{-7} \, \Omega \cdot \text{m} \), length \( L = 2.0 \, \text{m} \), and diameter \( d = 0.018 \, \text{m} \), we first calculate the cross-sectional area as \( A = \pi \left(\frac{d}{2}\right)^2 \). Plug these values into the resistance formula to find \( R_{\text{steel}} \).
• For the copper wire, use \( \rho_{\text{copper}} = 1.72 \times 10^{-8} \, \Omega \cdot \text{m} \), length \( L = 35 \, \text{m} \), and diameter \( d = 0.008 \, \text{m} \) to calculate the area and hence the resistance \( R_{\text{copper}} \).

The total resistance of the circuit is the sum of the resistances of both parts:\[ R_{\text{total}} = R_{\text{steel}} + R_{\text{copper}} \]This comprehensive approach ensures we accurately assess how much resistance the lightning rod-copper wire system provides against the freely flowing current.

Ohm's Law

Ohm's Law is a staple principle in the study of electric circuits, stating that the current through a conductor between two points is directly proportional to the voltage across the two points. It's expressed as:\[ V = I \times R \]where:

• \( V \) is the potential difference (voltage),
• \( I \) is the current through the conductor,
• \( R \) is the resistance.

In the problem at hand, we use Ohm's Law to determine the potential difference across the steel rod and copper wire during the burst of a lightning strike. Given:

• The current \( I = 15000 \, \text{A} \)
• And the total resistance \( R_{\text{total}} \) calculated from both the steel rod and copper wire

we substitute these values into Ohm's Law to find \( V \). This potential difference reflects the voltage drop that occurs due to the resistance encountered by the current as it travels through the materials. Understanding and applying Ohm's Law allows students to translate the abstract idea of voltage drop into tangible computation.

Energy Deposition

Energy deposition in circuits describes how electrical energy transforms and allocates throughout a system. This is quantifiable through the equation:\[ E = I^2 \times R \times t \]where:

• \( E \) is the energy (in joules),
• \( I \) is the current,
• \( R \) is the resistance,
• \( t \) is the time for which the current flows.

In the context of the exercise, calculate how much energy the lightning bolt transfers into the steel rod and copper wire over the short duration of 65 microseconds.

• First, find \( I^2 \) which scales the resistance's effect as current flows.
• Multiply with the total resistance \( R_{\text{total}} \) and the time \( t = 65 \times 10^{-6} \, \text{s} \).

This computation illustrates how energy disperses throughout the system as heat, causing the increase in thermal energy in the materials. Conceptualizing energy deposition provides insight into how electrical energy powers, heats, or impacts the structures within a circuit.