A cylindrical wire of radius 0.5 mm and conductivity `5 xx 10^7` S/m is subjected to an electric field of 10 mV/m. The expected value of current in the wire will be `x^3pi` mA. The value of x is ____ .

Answer

To find the expected value of current in the cylindrical wire, we can use the formula for current density and the relationship between current, conductivity, and electric field.

### Step 1: Understand the given parameters
- Radius of the wire, \( r = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \)
- Conductivity, \( \sigma = 5 \times 10^{7} \, \text{S/m} \)
- Electric field, \( E = 10 \, \text{mV/m} = 10 \times 10^{-3} \, \text{V/m} = 0.01 \, \text{V/m} \)

### Step 2: Calculate the current density (J)
The current density \( J \) can be calculated using the formula:
\[ J = \sigma E \]
Substituting the values:
\[ J = (5 \times 10^{7} \, \text{S/m})(0.01 \, \text{V/m}) = 5 \times 10^{5} \, \text{A/m}^2 \]

### Step 3: Calculate the cross-sectional area (A) of the wire
The cross-sectional area \( A \) of the cylindrical wire can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (5 \times 10^{-4})^2 = \pi (25 \times 10^{-8}) = 25\pi \times 10^{-8} \, \text{m}^2 \]

### Step 4: Calculate the total current (I)
The total current \( I \) can be calculated using the formula:
\[ I = J \cdot A \]
Substituting the values:
\[ I = (5 \times 10^{5} \, \text{A/m}^2) \cdot (25\pi \times 10^{-8} \, \text{m}^2) \]
\[ I = 125\pi \times 10^{-3} \, \text{A} = 125\pi \, \text{mA} \]

### Step 5: Relate the current to the given form \( x^3 \pi \)
According to the problem, the current is given as \( x^3 \pi \, \text{mA} \). Therefore, we can equate:
\[ 125\pi = x^3 \pi \]
Dividing both sides by \( \pi \):
\[ 125 = x^3 \]

### Step 6: Solve for \( x \)
Taking the cube root of both sides:
\[ x = \sqrt[3]{125} = 5 \]

Thus, the value of \( x \) is **5**.