Consider a resistor of uniform cross-sectional area connected to a battery of internal resistance zero. If the length of the resistor is doubled by stretching it, then

Options

A. the current will become four times
B. the electric field in the wire will become half
C. the thermal power produced by the resistor will become one fourth
D. The product of the current density and conductance will become half.

Explanation

To solve the problem, we need to analyze the effects of doubling the length of a resistor while keeping its volume constant. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between resistance, length, and area
The resistance \( R \) of a uniform resistor is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the resistor,
- \( A \) is the cross-sectional area.

### Step 2: Determine the new dimensions after stretching
When the length of the resistor is doubled, we have:
\[ L' = 2L \]
Since the volume \( V \) of the resistor remains constant, we can express the volume as:
\[ V = A \cdot L \]
After stretching, the new volume is:
\[ V' = A' \cdot L' = A' \cdot (2L) \]
Setting the volumes equal gives:
\[ A \cdot L = A' \cdot (2L) \]
From this, we can solve for the new area \( A' \):
\[ A' = \frac{A}{2} \]

### Step 3: Calculate the new resistance
Now we can substitute \( L' \) and \( A' \) into the resistance formula:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\frac{A}{2}} = \frac{2\rho L \cdot 2}{A} = \frac{4\rho L}{A} = 4R \]
Thus, the new resistance \( R' \) is four times the original resistance \( R \).

### Step 4: Analyze the current through the resistor
Using Ohm's law, the current \( I \) through the resistor is given by:
\[ I = \frac{E}{R} \]
where \( E \) is the electromotive force (emf) of the battery. The new current \( I' \) through the modified resistor is:
\[ I' = \frac{E}{R'} = \frac{E}{4R} = \frac{I}{4} \]
This indicates that the new current is one-fourth of the original current.

### Step 5: Determine the electric field in the wire
The electric field \( E \) in the wire can be expressed as:
\[ E = \frac{V}{L} \]
For the new wire, the electric field \( E' \) becomes:
\[ E' = \frac{V}{L'} = \frac{V}{2L} = \frac{E}{2} \]
Thus, the electric field in the wire is halved.

### Step 6: Calculate the thermal power produced by the resistor
The power \( P \) dissipated in the resistor is given by:
\[ P = \frac{V^2}{R} \]
For the new resistor, the power \( P' \) is:
\[ P' = \frac{V^2}{R'} = \frac{V^2}{4R} = \frac{P}{4} \]
This shows that the thermal power produced by the resistor is also one-fourth of the original power.

### Step 7: Analyze the product of current density and conductance
The current density \( J \) is given by:
\[ J = \frac{I}{A} \]
And the conductance \( G \) is the reciprocal of resistance:
\[ G = \frac{1}{R} \]
The product of current density and conductance \( JG \) is:
\[ JG = \frac{I}{A} \cdot \frac{1}{R} \]
Since both current and area have changed, we need to analyze how they affect the product. However, since the resistivity and geometry have changed in a way that keeps the product constant, we conclude that \( JG \) remains unchanged.

### Summary of Results
1. The new resistance \( R' = 4R \).
2. The new current \( I' = \frac{I}{4} \).
3. The electric field \( E' = \frac{E}{2} \).
4. The thermal power \( P' = \frac{P}{4} \).
5. The product of current density and conductance remains constant.