Animated Solution for Physics - Current Electricity: Two bars of radius $r$ and $2r$ are kept in contact as shown. An electric current $I$ is passed through the bars. Which one of following is correct ?

A Tale of Two Cylinders

Imagine you are observing a steady flow of water through a pipe that suddenly narrows down. The amount of water entering the wide section must equal the amount of water exiting the narrow section. In the electrical world, this is the essence of a series circuit.

We are given two cylindrical bars, $AB$ and $BC$, joined end-to-end. Because they are connected in series, the electric current $I$ flowing through them is absolutely identical. There are no alternate paths for the electrons to take, so the rate of flow of charge remains constant throughout the entire structure.

To determine which bar produces more heat, we first need to understand their electrical resistance. The resistance of a cylindrical conductor is given by the formula:

Let's calculate the resistance for each section individually.

For the thicker bar $AB$, the length is $l/2$ and the radius is $2r$. Plugging these into our formula, we get:

For the thinner bar $BC$, the length is also $l/2$, but the radius is just $r$. Its resistance is:

If we take the ratio of these two resistances, we find something very interesting:

This tells us that the thinner bar $BC$ has exactly four times the resistance of the thicker bar $AB$.

Now, let's bring in Joule's law of heating, which states that the heat produced in a conductor is:

Since the current $I$ and the time $t$ are the same for both bars, the heat produced is directly proportional to the resistance ($H\propto R$). Because $R_{BC}=4R_{AB}$, it naturally follows that:

Therefore, the heat produced in bar $BC$ is 4 times the heat produced in bar $AB$.

Why Other Options Fail

It is always a good practice to analyze why the other options are incorrect. Let's look at the other physical quantities:

1. Current Density ($J$): Defined as current per unit area ($J=\frac{I}{A}$). Since the area of $BC$ is one-fourth that of $AB$, its current density is four times higher ($J_{BC}=4J_{AB}$). 2. Electric Field ($E$): From the microscopic form of Ohm's law, $E=\rho J$. Since $J$ is four times higher in $BC$, the electric field is also four times higher ($E_{BC}=4E_{AB}$). 3. Potential Difference ($V$): Using Ohm's law ($V=IR$), since the resistance of $BC$ is four times higher, the potential drop across it is also four times higher ($V_{BC}=4V_{AB}$).

By systematically breaking down the geometry and applying fundamental laws, we can confidently conclude that only the statement regarding heat production is correct.