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Question

A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T, if the volume charge density of the rod is $\rho$, how long will the charges take to travel a distance d?

The correct answer is
$\frac{\rho dA}{I}$

Solving for Charge Travel Time in a Copper Rod

This section provides a step-by-step derivation to find the time ($t$) charges take to travel a distance ($d$) through a copper rod carrying a uniform current ($I$).

Relevant Physics Formulas

We utilize the following fundamental physics principles:

  • The definition of electric current: $I = \frac{Q}{t}$, where $I$ is the current, $Q$ is the amount of charge, and $t$ is the time taken for the charge to flow.
  • The relationship between total charge, volume charge density, and volume: $Q = \rho \times V$, where $\rho$ is the volume charge density and $V$ is the volume occupied by the charge.
  • The volume of a cylindrical segment (like a portion of the rod): $V = A \times d$, where $A$ is the cross-sectional area and $d$ is the length or distance.

The temperature $T$ mentioned in the question is not relevant for calculating the time taken for charge travel based on the given parameters.

Calculation Steps

  1. Determine the volume ($V$) of the section of the copper rod corresponding to the distance $d$. Given the cross-sectional area $A$, the volume is: $V = A \times d$
  2. Calculate the total charge ($Q$) contained within this volume. Using the provided volume charge density ($\rho$): $Q = \rho \times V = \rho \times A \times d$
  3. Calculate the time ($t$) required for this charge $Q$ to flow, using the definition of current $I = \frac{Q}{t}$. Substitute the expression for $Q$ derived in the previous step: $I = \frac{\rho A d}{t}$ Rearrange the formula to solve for the time ($t$): $t = \frac{\rho A d}{I}$

The calculated time it takes for the charges to travel the distance $d$ is $t = \frac{\rho A d}{I}$.

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Similar Questions

  1. A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3} \text{ J}$. The value of $\alpha$ is ______.
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Important Questions from Electricity and Magnetism

  1. A short bar magnet placed with its axis at $30^\circ$ with an external field of 800 Gauss, experiences a torque of 0.016 N.m. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3} \text{ J}$. The value of $\alpha$ is ______.
  2. An electromagnetic wave of frequency $100\text{ MHz}$ propagates through a medium of conductivity, $\sigma = 10\text{ mho/m}$. The ratio of maximum conduction current density to maximum displacement current density is __________.
    $\left[ \text{Take } \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ Nm}^2\text{/C}^2 \right]$
  3. An inductor stores $16\text{ J}$ of magnetic field energy and dissipates $32\text{ W}$ of thermal energy due to its resistance when an a.c. current of $2\text{ A}$ (rms) and frequency $50\text{ Hz}$ flows through it. The ratio of inductive reactance to its resistance is _________. ($\pi = 3.14$)
  4. A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of $60^\circ$ with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ___________ mV. (Take $g = 10\text{ m/s}^2$)
  5. Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for $V = 5\sin(100t)$ The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is ___________ $\mu\text{F}$. 

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