A rectangular block of metal of resistivity `p` has dimensions `d xx 2d xx 3d`. A potential difference `V` is applied between two opposite faces of the block.
(a) To which two faces of the block should the potential difference `V` be applied to give the maximum current density? What is the maximum current density?
(b) To which two faces of the block should the potential difference `V` be applied to give the maximum current? What is this maximum current?

Explanation

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): Maximum Current Density

1. **Understanding Current Density**:
The current density \( J \) is defined as:
\[ J = \frac{I}{A} \]
where \( I \) is the current and \( A \) is the cross-sectional area through which the current flows.

2. **Using Ohm's Law**:
According to Ohm's Law, the current \( I \) can be expressed as:
\[ I = \frac{V}{R} \]
where \( V \) is the potential difference and \( R \) is the resistance.

3. **Resistance of the Block**:
The resistance \( R \) of the block can be calculated using:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity, \( L \) is the length of the block in the direction of current flow, and \( A \) is the cross-sectional area.

4. **Substituting Resistance into Current Density**:
Substituting \( R \) into the equation for current density:
\[ J = \frac{I}{A} = \frac{V}{R \cdot A} = \frac{V \cdot A}{\rho L} \]

5. **Finding the Dimensions**:
The dimensions of the block are \( d \times 2d \times 3d \). Depending on which faces the potential difference \( V \) is applied, the length \( L \) will vary:
- If \( V \) is applied across the faces of area \( 2d \times 3d \), then \( L = d \).
- If \( V \) is applied across the faces of area \( d \times 3d \), then \( L = 2d \).
- If \( V \) is applied across the faces of area \( d \times 2d \), then \( L = 3d \).

6. **Maximizing Current Density**:
Since \( J \) is inversely proportional to \( L \), to maximize \( J \), we need to minimize \( L \). The minimum length \( L \) is \( d \).

7. **Calculating Maximum Current Density**:
Using \( L = d \):
\[ J_{\text{max}} = \frac{V}{\rho d} \]
Therefore, the potential difference should be applied across the faces of area \( 2d \times 3d \).

### Part (b): Maximum Current

1. **Using Current Formula**:
From Ohm's Law, we have:
\[ I = \frac{V}{R} \]
Substituting for \( R \):
\[ I = \frac{V \cdot A}{\rho L} \]

2. **Maximizing Current**:
To maximize \( I \), we need to maximize the area \( A \) and minimize the length \( L \).
- The maximum area \( A \) is \( 2d \times 3d = 6d^2 \).
- The minimum length \( L \) is \( d \).

3. **Calculating Maximum Current**:
Substituting these values into the equation:
\[ I_{\text{max}} = \frac{V \cdot (2d \cdot 3d)}{\rho d} = \frac{6Vd^2}{\rho d} = \frac{6Vd}{\rho} \]

### Summary of Solutions:
- **(a)** The potential difference \( V \) should be applied across the faces of area \( 2d \times 3d \) for maximum current density, which is:
\[ J_{\text{max}} = \frac{V}{\rho d} \]

- **(b)** The potential difference \( V \) should be applied across the faces of area \( 2d \times 3d \) for maximum current, which is:
\[ I_{\text{max}} = \frac{6Vd}{\rho} \]